Prove that $sqrt[8]5 > sqrt[9]6 > sqrt[10]7 > cdots$












2












$begingroup$



Prove that $sqrt[8]5 > sqrt[9]6 > sqrt[10]7 > cdots$




My friend came up with this and gave this to me as a challenge and I'm totally stuck.



I have tried proving this by induction $root{n+3}of{n} > root{n+4} of {n+1} $ for all integers $n geq 5$ with no luck. I don't even know how to prove the base case without a calculator. Also, it turns out that this is not true for $n leq 4$. Why would this inequality only true from $5$ onwards?






inequality radicals







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    $$sqrt[x+3]x $$ is a decreasing function
    $endgroup$
    – lab bhattacharjee
    Dec 27 '18 at 7:47






  • 1




    $begingroup$
    @labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
    $endgroup$
    – Matti P.
    Dec 27 '18 at 7:53






  • 1




    $begingroup$
    See this thread for cues. As you tagged this calculus presumably using derivatives is allowed, so you can look at Yves Daoust's answer in that thread in particular.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:08






  • 1




    $begingroup$
    Anyway $$Dx^{1/(x+3)}=-frac{x^{frac{1}{x+3}-1} (-x+x log (x)-3)}{(x+3)^2}.$$ Looking at that form it is clear which factor determines the sign.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:08








  • 2




    $begingroup$
    On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ is rather close to zero. Anyway, the dervative will take care of the infinite tail of the inequalities.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:24


















2












$begingroup$



Prove that $sqrt[8]5 > sqrt[9]6 > sqrt[10]7 > cdots$




My friend came up with this and gave this to me as a challenge and I'm totally stuck.



I have tried proving this by induction $root{n+3}of{n} > root{n+4} of {n+1} $ for all integers $n geq 5$ with no luck. I don't even know how to prove the base case without a calculator. Also, it turns out that this is not true for $n leq 4$. Why would this inequality only true from $5$ onwards?






inequality radicals







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    $$sqrt[x+3]x $$ is a decreasing function
    $endgroup$
    – lab bhattacharjee
    Dec 27 '18 at 7:47






  • 1




    $begingroup$
    @labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
    $endgroup$
    – Matti P.
    Dec 27 '18 at 7:53






  • 1




    $begingroup$
    See this thread for cues. As you tagged this calculus presumably using derivatives is allowed, so you can look at Yves Daoust's answer in that thread in particular.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:08






  • 1




    $begingroup$
    Anyway $$Dx^{1/(x+3)}=-frac{x^{frac{1}{x+3}-1} (-x+x log (x)-3)}{(x+3)^2}.$$ Looking at that form it is clear which factor determines the sign.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:08








  • 2




    $begingroup$
    On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ is rather close to zero. Anyway, the dervative will take care of the infinite tail of the inequalities.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:24
















2












2








2


2



$begingroup$



Prove that $sqrt[8]5 > sqrt[9]6 > sqrt[10]7 > cdots$




My friend came up with this and gave this to me as a challenge and I'm totally stuck.



I have tried proving this by induction $root{n+3}of{n} > root{n+4} of {n+1} $ for all integers $n geq 5$ with no luck. I don't even know how to prove the base case without a calculator. Also, it turns out that this is not true for $n leq 4$. Why would this inequality only true from $5$ onwards?






inequality radicals







share|cite|improve this question









$endgroup$





Prove that $sqrt[8]5 > sqrt[9]6 > sqrt[10]7 > cdots$




My friend came up with this and gave this to me as a challenge and I'm totally stuck.



I have tried proving this by induction $root{n+3}of{n} > root{n+4} of {n+1} $ for all integers $n geq 5$ with no luck. I don't even know how to prove the base case without a calculator. Also, it turns out that this is not true for $n leq 4$. Why would this inequality only true from $5$ onwards?






inequality radicals




inequality radicals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 27 '18 at 7:43









MintMint

5311417




5311417








  • 6




    $begingroup$
    $$sqrt[x+3]x $$ is a decreasing function
    $endgroup$
    – lab bhattacharjee
    Dec 27 '18 at 7:47






  • 1




    $begingroup$
    @labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
    $endgroup$
    – Matti P.
    Dec 27 '18 at 7:53






  • 1




    $begingroup$
    See this thread for cues. As you tagged this calculus presumably using derivatives is allowed, so you can look at Yves Daoust's answer in that thread in particular.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:08






  • 1




    $begingroup$
    Anyway $$Dx^{1/(x+3)}=-frac{x^{frac{1}{x+3}-1} (-x+x log (x)-3)}{(x+3)^2}.$$ Looking at that form it is clear which factor determines the sign.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:08








  • 2




    $begingroup$
    On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ is rather close to zero. Anyway, the dervative will take care of the infinite tail of the inequalities.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:24
















  • 6




    $begingroup$
    $$sqrt[x+3]x $$ is a decreasing function
    $endgroup$
    – lab bhattacharjee
    Dec 27 '18 at 7:47






  • 1




    $begingroup$
    @labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
    $endgroup$
    – Matti P.
    Dec 27 '18 at 7:53






  • 1




    $begingroup$
    See this thread for cues. As you tagged this calculus presumably using derivatives is allowed, so you can look at Yves Daoust's answer in that thread in particular.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:08






  • 1




    $begingroup$
    Anyway $$Dx^{1/(x+3)}=-frac{x^{frac{1}{x+3}-1} (-x+x log (x)-3)}{(x+3)^2}.$$ Looking at that form it is clear which factor determines the sign.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:08








  • 2




    $begingroup$
    On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ is rather close to zero. Anyway, the dervative will take care of the infinite tail of the inequalities.
    $endgroup$
    – Jyrki Lahtonen
    Dec 27 '18 at 8:24










6




6




$begingroup$
$$sqrt[x+3]x $$ is a decreasing function
$endgroup$
– lab bhattacharjee
Dec 27 '18 at 7:47




$begingroup$
$$sqrt[x+3]x $$ is a decreasing function
$endgroup$
– lab bhattacharjee
Dec 27 '18 at 7:47




1




1




$begingroup$
@labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
$endgroup$
– Matti P.
Dec 27 '18 at 7:53




$begingroup$
@labbhattacharjee gives a good hint. Just analyse $f(x) = sqrt[x+3]{x}$.
$endgroup$
– Matti P.
Dec 27 '18 at 7:53




1




1




$begingroup$
See this thread for cues. As you tagged this calculus presumably using derivatives is allowed, so you can look at Yves Daoust's answer in that thread in particular.
$endgroup$
– Jyrki Lahtonen
Dec 27 '18 at 8:08




$begingroup$
See this thread for cues. As you tagged this calculus presumably using derivatives is allowed, so you can look at Yves Daoust's answer in that thread in particular.
$endgroup$
– Jyrki Lahtonen
Dec 27 '18 at 8:08




1




1




$begingroup$
Anyway $$Dx^{1/(x+3)}=-frac{x^{frac{1}{x+3}-1} (-x+x log (x)-3)}{(x+3)^2}.$$ Looking at that form it is clear which factor determines the sign.
$endgroup$
– Jyrki Lahtonen
Dec 27 '18 at 8:08






$begingroup$
Anyway $$Dx^{1/(x+3)}=-frac{x^{frac{1}{x+3}-1} (-x+x log (x)-3)}{(x+3)^2}.$$ Looking at that form it is clear which factor determines the sign.
$endgroup$
– Jyrki Lahtonen
Dec 27 '18 at 8:08






2




2




$begingroup$
On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ is rather close to zero. Anyway, the dervative will take care of the infinite tail of the inequalities.
$endgroup$
– Jyrki Lahtonen
Dec 27 '18 at 8:24






$begingroup$
On second thought, you do need careful estimates to prove that the derivative above is negative already at $x=5$. After all, $5ln 5-8$ is rather close to zero. Anyway, the dervative will take care of the infinite tail of the inequalities.
$endgroup$
– Jyrki Lahtonen
Dec 27 '18 at 8:24












3 Answers
3






active

oldest

votes


















2












$begingroup$

Let $f(x)=frac {1}{x+3}ln x .$ Then $f'(x)=frac {x+3-xln x}{x(x+3)}.$



Let $g(x)=x+3-xln x.$ Then $g'(x)=-ln x.$ Now $g(x)$ is strictly decreasing for $xgeq 5$ because $g'(x)<0$ for $xgeq 5.$ By calculation $g(5)<0.$ So $g(x)leq g(5)<0$ for $xgeq 5.$



Therefore $f'(x)=frac {g(x)}{x(x+3)}<0$ for $xgeq 5,$ so $f(x)$ is strictly decreasing for $xgeq 5.$ So $e^{f(x)}=x^{1/(x+3)}$ is strictly decreasing for $xge 5.$



Remark: $g(x)$ is strictly decreasing for $xgeq 1$ but for small $x>1$ we have $g(x)>0$. E.g. $g(4)>0$. And $5$ is the least $nin Bbb N$ such that $g(n)<0$, i.e. such that $f'(n)<0.$



Remark. $g(5)<0iff e^8<5^5.$ We have $e<2.72implies e^2<7.3984<7.4implies$ $implies e^4<7.4^2=54.76<55implies$
$e^8<55^2=3025<3125=5^5.$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    We can also obtain $g(5)<0$ by knowing that $ln 10>2.3$ and $ln 2<0.7,$ as $g(5)=8-5(ln 10-ln 2)$.
    $endgroup$
    – DanielWainfleet
    Dec 27 '18 at 8:44



















2












$begingroup$

What yo need is $n^{frac 1 {n+3}} >(n+1)^{frac 1 {n+4}}$ which is same as $n^{n+4} >(n+1)^{n+3}$ or $(1+frac 1 n)^{n+3} <n$. In other words you need $(n+3)ln , (1+frac 1 n) <ln, n$Since $ln , (1+frac 1 n) <frac 1 n$ it is enough to show that $frac {n+3} n <ln, n$ or $1+frac 3 n <ln, n$. Since we have $n geq 5$ we have $1+frac 3 n leq frac 8 5 < ln 5 leq ln , n$ . [$e^{1.6} =4.965302...$].






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    For $ngeq5$ we need to prove that $$n^{frac{1}{n+3}}>(n+1)^{frac{1}{n+4}}$$ or
    $$n^{n+4}>(n+1)^{n+3}$$ or
    $$frac{n^4}{(n+1)^3}>left(1+frac{1}{n}right)^n$$ and since $left(1+frac{1}{n}right)^n<e,$ it's enough to prove that
    $$n^4-e(n+1)^3>0.$$
    But the polynomial $n^4-e(n+1)^3$ has one changing of coefficients signs.



    Thus, by the Descartes's rule this polynomial has unique positive root.



    Id est, it remains to check that
    $$frac{5^4}{6^3}>e$$ and since $e=2.718...<2.75,$ it's enough to prove that
    $$frac{625}{216}>2.75$$ or $$625>594.$$
    Done!






    share|cite|improve this answer











    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053683%2fprove-that-sqrt85-sqrt96-sqrt107-cdots%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Let $f(x)=frac {1}{x+3}ln x .$ Then $f'(x)=frac {x+3-xln x}{x(x+3)}.$



      Let $g(x)=x+3-xln x.$ Then $g'(x)=-ln x.$ Now $g(x)$ is strictly decreasing for $xgeq 5$ because $g'(x)<0$ for $xgeq 5.$ By calculation $g(5)<0.$ So $g(x)leq g(5)<0$ for $xgeq 5.$



      Therefore $f'(x)=frac {g(x)}{x(x+3)}<0$ for $xgeq 5,$ so $f(x)$ is strictly decreasing for $xgeq 5.$ So $e^{f(x)}=x^{1/(x+3)}$ is strictly decreasing for $xge 5.$



      Remark: $g(x)$ is strictly decreasing for $xgeq 1$ but for small $x>1$ we have $g(x)>0$. E.g. $g(4)>0$. And $5$ is the least $nin Bbb N$ such that $g(n)<0$, i.e. such that $f'(n)<0.$



      Remark. $g(5)<0iff e^8<5^5.$ We have $e<2.72implies e^2<7.3984<7.4implies$ $implies e^4<7.4^2=54.76<55implies$
      $e^8<55^2=3025<3125=5^5.$






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        We can also obtain $g(5)<0$ by knowing that $ln 10>2.3$ and $ln 2<0.7,$ as $g(5)=8-5(ln 10-ln 2)$.
        $endgroup$
        – DanielWainfleet
        Dec 27 '18 at 8:44
















      2












      $begingroup$

      Let $f(x)=frac {1}{x+3}ln x .$ Then $f'(x)=frac {x+3-xln x}{x(x+3)}.$



      Let $g(x)=x+3-xln x.$ Then $g'(x)=-ln x.$ Now $g(x)$ is strictly decreasing for $xgeq 5$ because $g'(x)<0$ for $xgeq 5.$ By calculation $g(5)<0.$ So $g(x)leq g(5)<0$ for $xgeq 5.$



      Therefore $f'(x)=frac {g(x)}{x(x+3)}<0$ for $xgeq 5,$ so $f(x)$ is strictly decreasing for $xgeq 5.$ So $e^{f(x)}=x^{1/(x+3)}$ is strictly decreasing for $xge 5.$



      Remark: $g(x)$ is strictly decreasing for $xgeq 1$ but for small $x>1$ we have $g(x)>0$. E.g. $g(4)>0$. And $5$ is the least $nin Bbb N$ such that $g(n)<0$, i.e. such that $f'(n)<0.$



      Remark. $g(5)<0iff e^8<5^5.$ We have $e<2.72implies e^2<7.3984<7.4implies$ $implies e^4<7.4^2=54.76<55implies$
      $e^8<55^2=3025<3125=5^5.$






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        We can also obtain $g(5)<0$ by knowing that $ln 10>2.3$ and $ln 2<0.7,$ as $g(5)=8-5(ln 10-ln 2)$.
        $endgroup$
        – DanielWainfleet
        Dec 27 '18 at 8:44














      2












      2








      2





      $begingroup$

      Let $f(x)=frac {1}{x+3}ln x .$ Then $f'(x)=frac {x+3-xln x}{x(x+3)}.$



      Let $g(x)=x+3-xln x.$ Then $g'(x)=-ln x.$ Now $g(x)$ is strictly decreasing for $xgeq 5$ because $g'(x)<0$ for $xgeq 5.$ By calculation $g(5)<0.$ So $g(x)leq g(5)<0$ for $xgeq 5.$



      Therefore $f'(x)=frac {g(x)}{x(x+3)}<0$ for $xgeq 5,$ so $f(x)$ is strictly decreasing for $xgeq 5.$ So $e^{f(x)}=x^{1/(x+3)}$ is strictly decreasing for $xge 5.$



      Remark: $g(x)$ is strictly decreasing for $xgeq 1$ but for small $x>1$ we have $g(x)>0$. E.g. $g(4)>0$. And $5$ is the least $nin Bbb N$ such that $g(n)<0$, i.e. such that $f'(n)<0.$



      Remark. $g(5)<0iff e^8<5^5.$ We have $e<2.72implies e^2<7.3984<7.4implies$ $implies e^4<7.4^2=54.76<55implies$
      $e^8<55^2=3025<3125=5^5.$






      share|cite|improve this answer











      $endgroup$



      Let $f(x)=frac {1}{x+3}ln x .$ Then $f'(x)=frac {x+3-xln x}{x(x+3)}.$



      Let $g(x)=x+3-xln x.$ Then $g'(x)=-ln x.$ Now $g(x)$ is strictly decreasing for $xgeq 5$ because $g'(x)<0$ for $xgeq 5.$ By calculation $g(5)<0.$ So $g(x)leq g(5)<0$ for $xgeq 5.$



      Therefore $f'(x)=frac {g(x)}{x(x+3)}<0$ for $xgeq 5,$ so $f(x)$ is strictly decreasing for $xgeq 5.$ So $e^{f(x)}=x^{1/(x+3)}$ is strictly decreasing for $xge 5.$



      Remark: $g(x)$ is strictly decreasing for $xgeq 1$ but for small $x>1$ we have $g(x)>0$. E.g. $g(4)>0$. And $5$ is the least $nin Bbb N$ such that $g(n)<0$, i.e. such that $f'(n)<0.$



      Remark. $g(5)<0iff e^8<5^5.$ We have $e<2.72implies e^2<7.3984<7.4implies$ $implies e^4<7.4^2=54.76<55implies$
      $e^8<55^2=3025<3125=5^5.$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 27 '18 at 8:38

























      answered Dec 27 '18 at 8:15









      DanielWainfleetDanielWainfleet

      35.2k31648




      35.2k31648








      • 1




        $begingroup$
        We can also obtain $g(5)<0$ by knowing that $ln 10>2.3$ and $ln 2<0.7,$ as $g(5)=8-5(ln 10-ln 2)$.
        $endgroup$
        – DanielWainfleet
        Dec 27 '18 at 8:44














      • 1




        $begingroup$
        We can also obtain $g(5)<0$ by knowing that $ln 10>2.3$ and $ln 2<0.7,$ as $g(5)=8-5(ln 10-ln 2)$.
        $endgroup$
        – DanielWainfleet
        Dec 27 '18 at 8:44








      1




      1




      $begingroup$
      We can also obtain $g(5)<0$ by knowing that $ln 10>2.3$ and $ln 2<0.7,$ as $g(5)=8-5(ln 10-ln 2)$.
      $endgroup$
      – DanielWainfleet
      Dec 27 '18 at 8:44




      $begingroup$
      We can also obtain $g(5)<0$ by knowing that $ln 10>2.3$ and $ln 2<0.7,$ as $g(5)=8-5(ln 10-ln 2)$.
      $endgroup$
      – DanielWainfleet
      Dec 27 '18 at 8:44











      2












      $begingroup$

      What yo need is $n^{frac 1 {n+3}} >(n+1)^{frac 1 {n+4}}$ which is same as $n^{n+4} >(n+1)^{n+3}$ or $(1+frac 1 n)^{n+3} <n$. In other words you need $(n+3)ln , (1+frac 1 n) <ln, n$Since $ln , (1+frac 1 n) <frac 1 n$ it is enough to show that $frac {n+3} n <ln, n$ or $1+frac 3 n <ln, n$. Since we have $n geq 5$ we have $1+frac 3 n leq frac 8 5 < ln 5 leq ln , n$ . [$e^{1.6} =4.965302...$].






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        What yo need is $n^{frac 1 {n+3}} >(n+1)^{frac 1 {n+4}}$ which is same as $n^{n+4} >(n+1)^{n+3}$ or $(1+frac 1 n)^{n+3} <n$. In other words you need $(n+3)ln , (1+frac 1 n) <ln, n$Since $ln , (1+frac 1 n) <frac 1 n$ it is enough to show that $frac {n+3} n <ln, n$ or $1+frac 3 n <ln, n$. Since we have $n geq 5$ we have $1+frac 3 n leq frac 8 5 < ln 5 leq ln , n$ . [$e^{1.6} =4.965302...$].






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          What yo need is $n^{frac 1 {n+3}} >(n+1)^{frac 1 {n+4}}$ which is same as $n^{n+4} >(n+1)^{n+3}$ or $(1+frac 1 n)^{n+3} <n$. In other words you need $(n+3)ln , (1+frac 1 n) <ln, n$Since $ln , (1+frac 1 n) <frac 1 n$ it is enough to show that $frac {n+3} n <ln, n$ or $1+frac 3 n <ln, n$. Since we have $n geq 5$ we have $1+frac 3 n leq frac 8 5 < ln 5 leq ln , n$ . [$e^{1.6} =4.965302...$].






          share|cite|improve this answer









          $endgroup$



          What yo need is $n^{frac 1 {n+3}} >(n+1)^{frac 1 {n+4}}$ which is same as $n^{n+4} >(n+1)^{n+3}$ or $(1+frac 1 n)^{n+3} <n$. In other words you need $(n+3)ln , (1+frac 1 n) <ln, n$Since $ln , (1+frac 1 n) <frac 1 n$ it is enough to show that $frac {n+3} n <ln, n$ or $1+frac 3 n <ln, n$. Since we have $n geq 5$ we have $1+frac 3 n leq frac 8 5 < ln 5 leq ln , n$ . [$e^{1.6} =4.965302...$].







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 7:53









          Kavi Rama MurthyKavi Rama Murthy

          62.5k42262




          62.5k42262























              1












              $begingroup$

              For $ngeq5$ we need to prove that $$n^{frac{1}{n+3}}>(n+1)^{frac{1}{n+4}}$$ or
              $$n^{n+4}>(n+1)^{n+3}$$ or
              $$frac{n^4}{(n+1)^3}>left(1+frac{1}{n}right)^n$$ and since $left(1+frac{1}{n}right)^n<e,$ it's enough to prove that
              $$n^4-e(n+1)^3>0.$$
              But the polynomial $n^4-e(n+1)^3$ has one changing of coefficients signs.



              Thus, by the Descartes's rule this polynomial has unique positive root.



              Id est, it remains to check that
              $$frac{5^4}{6^3}>e$$ and since $e=2.718...<2.75,$ it's enough to prove that
              $$frac{625}{216}>2.75$$ or $$625>594.$$
              Done!






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                For $ngeq5$ we need to prove that $$n^{frac{1}{n+3}}>(n+1)^{frac{1}{n+4}}$$ or
                $$n^{n+4}>(n+1)^{n+3}$$ or
                $$frac{n^4}{(n+1)^3}>left(1+frac{1}{n}right)^n$$ and since $left(1+frac{1}{n}right)^n<e,$ it's enough to prove that
                $$n^4-e(n+1)^3>0.$$
                But the polynomial $n^4-e(n+1)^3$ has one changing of coefficients signs.



                Thus, by the Descartes's rule this polynomial has unique positive root.



                Id est, it remains to check that
                $$frac{5^4}{6^3}>e$$ and since $e=2.718...<2.75,$ it's enough to prove that
                $$frac{625}{216}>2.75$$ or $$625>594.$$
                Done!






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  For $ngeq5$ we need to prove that $$n^{frac{1}{n+3}}>(n+1)^{frac{1}{n+4}}$$ or
                  $$n^{n+4}>(n+1)^{n+3}$$ or
                  $$frac{n^4}{(n+1)^3}>left(1+frac{1}{n}right)^n$$ and since $left(1+frac{1}{n}right)^n<e,$ it's enough to prove that
                  $$n^4-e(n+1)^3>0.$$
                  But the polynomial $n^4-e(n+1)^3$ has one changing of coefficients signs.



                  Thus, by the Descartes's rule this polynomial has unique positive root.



                  Id est, it remains to check that
                  $$frac{5^4}{6^3}>e$$ and since $e=2.718...<2.75,$ it's enough to prove that
                  $$frac{625}{216}>2.75$$ or $$625>594.$$
                  Done!






                  share|cite|improve this answer











                  $endgroup$



                  For $ngeq5$ we need to prove that $$n^{frac{1}{n+3}}>(n+1)^{frac{1}{n+4}}$$ or
                  $$n^{n+4}>(n+1)^{n+3}$$ or
                  $$frac{n^4}{(n+1)^3}>left(1+frac{1}{n}right)^n$$ and since $left(1+frac{1}{n}right)^n<e,$ it's enough to prove that
                  $$n^4-e(n+1)^3>0.$$
                  But the polynomial $n^4-e(n+1)^3$ has one changing of coefficients signs.



                  Thus, by the Descartes's rule this polynomial has unique positive root.



                  Id est, it remains to check that
                  $$frac{5^4}{6^3}>e$$ and since $e=2.718...<2.75,$ it's enough to prove that
                  $$frac{625}{216}>2.75$$ or $$625>594.$$
                  Done!







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 27 '18 at 10:12

























                  answered Dec 27 '18 at 10:04









                  Michael RozenbergMichael Rozenberg

                  105k1892198




                  105k1892198






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053683%2fprove-that-sqrt85-sqrt96-sqrt107-cdots%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Ellipse (mathématiques)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                      Read more