Find $limlimits_{n to infty} sumlimits_{k=1}^{infty}frac{1}{k^{2}sqrt[k]{n}}sin^{2}left(frac{n pi}{k}right)$












3












$begingroup$



Find $$limlimits_{n to infty} sumlimits_{k=1}^{infty}frac{1}{k^{2}sqrt[k]{n}}sin^{2}left(frac{n pi}{k}right)$$




This is the first time that I am operating with $lim_{nto infty}lim_{k to infty}$ so I am unsure. My first idea would be to look at:



$frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})$ where $n in mathbb N$ is constant.



$frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})leq frac{1}{k^{2}sqrt[k]{n}}leqfrac{1}{k^{2}sqrt{n}}$



and $sum_{k=1}^{infty}frac{1}{k^{2}sqrt{n}}=frac{1}{sqrt{n}}sum_{k=1}^{infty}frac{1}{k^{2}}$



and we know $sum_{k=1}^{infty}frac{1}{k^{2}} < infty$ and taking $n to infty$ we get



$lim_{nto infty}frac{1}{sqrt{n}}sum_{k=1}^{infty}frac{1}{k^{2}}=0=lim_{n to infty} sum_{k=1}^{infty}frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})$



I assume this is incorrect. Help/Corrections would be greatly appreciated.






real-analysis sequences-and-series limits







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  • $begingroup$
    $cdots leqslant dfrac 1{k^2 sqrt n}$ holds for $k geqslant 2$, so it should be $$ frac 1n + frac 1{sqrt n}sum_2^infty frac 1{k^2} xrightarrow{n to +infty} 0.$$
    $endgroup$
    – xbh
    Dec 3 '18 at 9:54
















3












$begingroup$



Find $$limlimits_{n to infty} sumlimits_{k=1}^{infty}frac{1}{k^{2}sqrt[k]{n}}sin^{2}left(frac{n pi}{k}right)$$




This is the first time that I am operating with $lim_{nto infty}lim_{k to infty}$ so I am unsure. My first idea would be to look at:



$frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})$ where $n in mathbb N$ is constant.



$frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})leq frac{1}{k^{2}sqrt[k]{n}}leqfrac{1}{k^{2}sqrt{n}}$



and $sum_{k=1}^{infty}frac{1}{k^{2}sqrt{n}}=frac{1}{sqrt{n}}sum_{k=1}^{infty}frac{1}{k^{2}}$



and we know $sum_{k=1}^{infty}frac{1}{k^{2}} < infty$ and taking $n to infty$ we get



$lim_{nto infty}frac{1}{sqrt{n}}sum_{k=1}^{infty}frac{1}{k^{2}}=0=lim_{n to infty} sum_{k=1}^{infty}frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})$



I assume this is incorrect. Help/Corrections would be greatly appreciated.






real-analysis sequences-and-series limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    $cdots leqslant dfrac 1{k^2 sqrt n}$ holds for $k geqslant 2$, so it should be $$ frac 1n + frac 1{sqrt n}sum_2^infty frac 1{k^2} xrightarrow{n to +infty} 0.$$
    $endgroup$
    – xbh
    Dec 3 '18 at 9:54














3












3








3


1



$begingroup$



Find $$limlimits_{n to infty} sumlimits_{k=1}^{infty}frac{1}{k^{2}sqrt[k]{n}}sin^{2}left(frac{n pi}{k}right)$$




This is the first time that I am operating with $lim_{nto infty}lim_{k to infty}$ so I am unsure. My first idea would be to look at:



$frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})$ where $n in mathbb N$ is constant.



$frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})leq frac{1}{k^{2}sqrt[k]{n}}leqfrac{1}{k^{2}sqrt{n}}$



and $sum_{k=1}^{infty}frac{1}{k^{2}sqrt{n}}=frac{1}{sqrt{n}}sum_{k=1}^{infty}frac{1}{k^{2}}$



and we know $sum_{k=1}^{infty}frac{1}{k^{2}} < infty$ and taking $n to infty$ we get



$lim_{nto infty}frac{1}{sqrt{n}}sum_{k=1}^{infty}frac{1}{k^{2}}=0=lim_{n to infty} sum_{k=1}^{infty}frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})$



I assume this is incorrect. Help/Corrections would be greatly appreciated.






real-analysis sequences-and-series limits







share|cite|improve this question











$endgroup$





Find $$limlimits_{n to infty} sumlimits_{k=1}^{infty}frac{1}{k^{2}sqrt[k]{n}}sin^{2}left(frac{n pi}{k}right)$$




This is the first time that I am operating with $lim_{nto infty}lim_{k to infty}$ so I am unsure. My first idea would be to look at:



$frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})$ where $n in mathbb N$ is constant.



$frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})leq frac{1}{k^{2}sqrt[k]{n}}leqfrac{1}{k^{2}sqrt{n}}$



and $sum_{k=1}^{infty}frac{1}{k^{2}sqrt{n}}=frac{1}{sqrt{n}}sum_{k=1}^{infty}frac{1}{k^{2}}$



and we know $sum_{k=1}^{infty}frac{1}{k^{2}} < infty$ and taking $n to infty$ we get



$lim_{nto infty}frac{1}{sqrt{n}}sum_{k=1}^{infty}frac{1}{k^{2}}=0=lim_{n to infty} sum_{k=1}^{infty}frac{1}{k^{2}sqrt[k]{n}}sin^{2}(frac{n pi}{k})$



I assume this is incorrect. Help/Corrections would be greatly appreciated.






real-analysis sequences-and-series limits




real-analysis sequences-and-series limits






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edited Dec 3 '18 at 11:02









Did

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asked Dec 3 '18 at 9:41









SABOYSABOY

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567311












  • $begingroup$
    $cdots leqslant dfrac 1{k^2 sqrt n}$ holds for $k geqslant 2$, so it should be $$ frac 1n + frac 1{sqrt n}sum_2^infty frac 1{k^2} xrightarrow{n to +infty} 0.$$
    $endgroup$
    – xbh
    Dec 3 '18 at 9:54


















  • $begingroup$
    $cdots leqslant dfrac 1{k^2 sqrt n}$ holds for $k geqslant 2$, so it should be $$ frac 1n + frac 1{sqrt n}sum_2^infty frac 1{k^2} xrightarrow{n to +infty} 0.$$
    $endgroup$
    – xbh
    Dec 3 '18 at 9:54
















$begingroup$
$cdots leqslant dfrac 1{k^2 sqrt n}$ holds for $k geqslant 2$, so it should be $$ frac 1n + frac 1{sqrt n}sum_2^infty frac 1{k^2} xrightarrow{n to +infty} 0.$$
$endgroup$
– xbh
Dec 3 '18 at 9:54




$begingroup$
$cdots leqslant dfrac 1{k^2 sqrt n}$ holds for $k geqslant 2$, so it should be $$ frac 1n + frac 1{sqrt n}sum_2^infty frac 1{k^2} xrightarrow{n to +infty} 0.$$
$endgroup$
– xbh
Dec 3 '18 at 9:54










1 Answer
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$begingroup$

In your manipulations there is a mistake: note that for $kge 2$



$$sqrt{n}gesqrt[k]{n}impliesfrac1{k^2sqrt[k]{n}}gefrac1{k^2sqrt n}tag1$$



A way to solve this limit is using the Weierstrass M-test and the properties of uniform convergence of series.



Note that for all $kinBbb N_{ge 1}$ and $xge 1$ it holds that $sqrt[k]{x}ge 1$, consequently



$$frac1{k^2}gefrac1{k^2sqrt[k]{n}}gefrac1{k^2sqrt[k]{n}}sin^2(n pi/k)tag2$$



Hence by the M-test the series $sum_{k=1}^infty f_k(x)$, for $f_k(x):=frac1{k^2sqrt[k]{x}}sin^2(x pi/k)$, converges absolutely and uniformly for $xge 1$, so we can exchange limit and summation sign to find that the limit that we want to evaluate is indeed zero.






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    $begingroup$

    In your manipulations there is a mistake: note that for $kge 2$



    $$sqrt{n}gesqrt[k]{n}impliesfrac1{k^2sqrt[k]{n}}gefrac1{k^2sqrt n}tag1$$



    A way to solve this limit is using the Weierstrass M-test and the properties of uniform convergence of series.



    Note that for all $kinBbb N_{ge 1}$ and $xge 1$ it holds that $sqrt[k]{x}ge 1$, consequently



    $$frac1{k^2}gefrac1{k^2sqrt[k]{n}}gefrac1{k^2sqrt[k]{n}}sin^2(n pi/k)tag2$$



    Hence by the M-test the series $sum_{k=1}^infty f_k(x)$, for $f_k(x):=frac1{k^2sqrt[k]{x}}sin^2(x pi/k)$, converges absolutely and uniformly for $xge 1$, so we can exchange limit and summation sign to find that the limit that we want to evaluate is indeed zero.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      In your manipulations there is a mistake: note that for $kge 2$



      $$sqrt{n}gesqrt[k]{n}impliesfrac1{k^2sqrt[k]{n}}gefrac1{k^2sqrt n}tag1$$



      A way to solve this limit is using the Weierstrass M-test and the properties of uniform convergence of series.



      Note that for all $kinBbb N_{ge 1}$ and $xge 1$ it holds that $sqrt[k]{x}ge 1$, consequently



      $$frac1{k^2}gefrac1{k^2sqrt[k]{n}}gefrac1{k^2sqrt[k]{n}}sin^2(n pi/k)tag2$$



      Hence by the M-test the series $sum_{k=1}^infty f_k(x)$, for $f_k(x):=frac1{k^2sqrt[k]{x}}sin^2(x pi/k)$, converges absolutely and uniformly for $xge 1$, so we can exchange limit and summation sign to find that the limit that we want to evaluate is indeed zero.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        In your manipulations there is a mistake: note that for $kge 2$



        $$sqrt{n}gesqrt[k]{n}impliesfrac1{k^2sqrt[k]{n}}gefrac1{k^2sqrt n}tag1$$



        A way to solve this limit is using the Weierstrass M-test and the properties of uniform convergence of series.



        Note that for all $kinBbb N_{ge 1}$ and $xge 1$ it holds that $sqrt[k]{x}ge 1$, consequently



        $$frac1{k^2}gefrac1{k^2sqrt[k]{n}}gefrac1{k^2sqrt[k]{n}}sin^2(n pi/k)tag2$$



        Hence by the M-test the series $sum_{k=1}^infty f_k(x)$, for $f_k(x):=frac1{k^2sqrt[k]{x}}sin^2(x pi/k)$, converges absolutely and uniformly for $xge 1$, so we can exchange limit and summation sign to find that the limit that we want to evaluate is indeed zero.






        share|cite|improve this answer











        $endgroup$



        In your manipulations there is a mistake: note that for $kge 2$



        $$sqrt{n}gesqrt[k]{n}impliesfrac1{k^2sqrt[k]{n}}gefrac1{k^2sqrt n}tag1$$



        A way to solve this limit is using the Weierstrass M-test and the properties of uniform convergence of series.



        Note that for all $kinBbb N_{ge 1}$ and $xge 1$ it holds that $sqrt[k]{x}ge 1$, consequently



        $$frac1{k^2}gefrac1{k^2sqrt[k]{n}}gefrac1{k^2sqrt[k]{n}}sin^2(n pi/k)tag2$$



        Hence by the M-test the series $sum_{k=1}^infty f_k(x)$, for $f_k(x):=frac1{k^2sqrt[k]{x}}sin^2(x pi/k)$, converges absolutely and uniformly for $xge 1$, so we can exchange limit and summation sign to find that the limit that we want to evaluate is indeed zero.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 11:13

























        answered Dec 3 '18 at 10:38









        MasacrosoMasacroso

        13k41746




        13k41746






























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