Solve $limlimits_{ntoinfty}sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$
up vote
6
down vote
favorite
I am having great problems in solving this:
$$limlimits_{ntoinfty}sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$$
I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I have to use the third binomial formula here, so my next step would be:
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$ so
$$a-b=frac{a^3-b^3}{a^2+ab+b^2}$$
I tried expanding it as well, which led to absolutely nothing. These are my writings to this:
limits roots
edited Nov 22 at 20:10
Robert Howard
1,9181822
asked Nov 22 at 19:28
SacredScout
487
add a comment |
up vote
6
down vote
favorite
I am having great problems in solving this:
$$limlimits_{ntoinfty}sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$$
I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I have to use the third binomial formula here, so my next step would be:
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$ so
$$a-b=frac{a^3-b^3}{a^2+ab+b^2}$$
I tried expanding it as well, which led to absolutely nothing. These are my writings to this:
limits roots
edited Nov 22 at 20:10
Robert Howard
1,9181822
asked Nov 22 at 19:28
SacredScout
487
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I am having great problems in solving this:
$$limlimits_{ntoinfty}sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$$
I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I have to use the third binomial formula here, so my next step would be:
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$ so
$$a-b=frac{a^3-b^3}{a^2+ab+b^2}$$
I tried expanding it as well, which led to absolutely nothing. These are my writings to this:
limits roots
edited Nov 22 at 20:10
Robert Howard
1,9181822
asked Nov 22 at 19:28
SacredScout
487
I am having great problems in solving this:
$$limlimits_{ntoinfty}sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$$
I am trying to solve this for hours, no solution in sight. I tried so many ways on my paper here, which all lead to nonsense or to nowhere. I concluded that I have to use the third binomial formula here, so my next step would be:
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$ so
$$a-b=frac{a^3-b^3}{a^2+ab+b^2}$$
I tried expanding it as well, which led to absolutely nothing. These are my writings to this:
limits roots
limits roots
edited Nov 22 at 20:10
Robert Howard
1,9181822
asked Nov 22 at 19:28
SacredScout
487
edited Nov 22 at 20:10
Robert Howard
1,9181822
asked Nov 22 at 19:28
SacredScout
487
edited Nov 22 at 20:10
Robert Howard
1,9181822
edited Nov 22 at 20:10
Robert Howard
1,9181822
edited Nov 22 at 20:10
Robert Howard
1,9181822
1,9181822
asked Nov 22 at 19:28
SacredScout
487
asked Nov 22 at 19:28
SacredScout
487
asked Nov 22 at 19:28
SacredScout
487
487
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
up vote
7
down vote
accepted
$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{n+sqrt{n}-n}{(n+sqrt{n})^{2/3}+(n+sqrt{n})^{1/3}n^{1/3}+n^{2/3}}leq frac{sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=frac{sqrt{n}}{3n^{2/3}}rightarrow 0$
answered Nov 22 at 19:36
John_Wick
1,199111
That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
– SacredScout
Nov 22 at 20:02
add a comment |
up vote
3
down vote
By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have
$$sqrt[3]{n+sqrt{n}}=sqrt[3]{n}, left(1+frac1{sqrt n}right)^frac13=sqrt[3]{n}+frac{sqrt[3]{n}}{3sqrt n}+oleft(frac{sqrt[3]{n}}{sqrt n}right)=sqrt[3]{n}+frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$
therefore
$$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$
edited Nov 22 at 23:45
amWhy
191k28224439
answered Nov 22 at 19:59
gimusi
92.8k94495
May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
– William McGonagall
Nov 27 at 7:08
1
@WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
– gimusi
Nov 27 at 7:14
Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
– William McGonagall
Nov 27 at 7:26
@WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
– gimusi
Nov 27 at 7:37
Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
– William McGonagall
Nov 27 at 7:42
add a comment |
up vote
3
down vote
Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$
We have that
$$0< sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{sqrt[3]{n}}{sqrt{n}}cdotfrac{
sqrt[3]{1+frac{1}{sqrt{n}}}-1}{frac{1}{sqrt{n}}}leq frac{1}{sqrt[6]{n}}cdot frac{1}{3}.$$
where we used the Bernoulli's inequality
$$(1+x)^rleq 1+rx$$
with $r=1/3in(0,1)$ and $x=1/sqrt{n}$.
Can you take it from here?
P.S. By using this approach we are also able to find
$$limlimits_{ntoinfty}left((n+sqrt{n})^r-n^rright)$$
for any $r<1/2$. Note that if we replace $sqrt[3]{ }$ with say $sqrt[5]{ }$ then the "algebraic way" could be very annoying!
answered Nov 22 at 19:30
Robert Z
92.4k1058129
That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
– SacredScout
Nov 22 at 19:37
1
Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
– Robert Z
Nov 22 at 19:39
I must have been blind..... Got that! I still have to admit that I don't get the denominator.
– SacredScout
Nov 22 at 19:45
$sqrt{n}cdot frac{1}{sqrt{n}}=1$...
– Robert Z
Nov 22 at 19:46
It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
– SacredScout
Nov 27 at 15:23
add a comment |
up vote
2
down vote
As you suggested,
begin{align}
sqrt[3]{n+sqrt n} - sqrt[3]{n} &= frac{n+sqrt n - n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}\
&= frac{sqrt n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}.
end{align}
The fastest growing term is evidently $n^{2/3}$. Can you finish?
answered Nov 22 at 19:31
MisterRiemann
5,7131624
add a comment |
up vote
0
down vote
Consider the function $f(x)=x^{1/3}$. By the mean value theorem there's a number $yin (n, n+sqrt n)$ such that
$$
f(n+sqrt n) - f(n) = f'(y)(n+sqrt n - n)= frac{y^{-2/3}}{3}sqrt n<n^{-2/3}sqrt n=n^{-1/6}to 0.
$$
answered Nov 24 at 8:21
User
483514
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{n+sqrt{n}-n}{(n+sqrt{n})^{2/3}+(n+sqrt{n})^{1/3}n^{1/3}+n^{2/3}}leq frac{sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=frac{sqrt{n}}{3n^{2/3}}rightarrow 0$
answered Nov 22 at 19:36
John_Wick
1,199111
That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
– SacredScout
Nov 22 at 20:02
add a comment |
up vote
7
down vote
accepted
$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{n+sqrt{n}-n}{(n+sqrt{n})^{2/3}+(n+sqrt{n})^{1/3}n^{1/3}+n^{2/3}}leq frac{sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=frac{sqrt{n}}{3n^{2/3}}rightarrow 0$
answered Nov 22 at 19:36
John_Wick
1,199111
That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
– SacredScout
Nov 22 at 20:02
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{n+sqrt{n}-n}{(n+sqrt{n})^{2/3}+(n+sqrt{n})^{1/3}n^{1/3}+n^{2/3}}leq frac{sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=frac{sqrt{n}}{3n^{2/3}}rightarrow 0$
answered Nov 22 at 19:36
John_Wick
1,199111
$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{n+sqrt{n}-n}{(n+sqrt{n})^{2/3}+(n+sqrt{n})^{1/3}n^{1/3}+n^{2/3}}leq frac{sqrt{n}}{n^{2/3}+n^{1/3}n^{1/3}+n^{2/3}}=frac{sqrt{n}}{3n^{2/3}}rightarrow 0$
answered Nov 22 at 19:36
John_Wick
1,199111
answered Nov 22 at 19:36
John_Wick
1,199111
answered Nov 22 at 19:36
John_Wick
1,199111
answered Nov 22 at 19:36
John_Wick
1,199111
1,199111
That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
– SacredScout
Nov 22 at 20:02
add a comment |
That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
– SacredScout
Nov 22 at 20:02
That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
– SacredScout
Nov 22 at 20:02
That is the clearest explanation I got, I can really understand everything there - which I can seldom say because my knowledge of maths is very limited (for now). I have chosen yours as the best, thank you! Keep your service up!
– SacredScout
Nov 22 at 20:02
add a comment |
up vote
3
down vote
By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have
$$sqrt[3]{n+sqrt{n}}=sqrt[3]{n}, left(1+frac1{sqrt n}right)^frac13=sqrt[3]{n}+frac{sqrt[3]{n}}{3sqrt n}+oleft(frac{sqrt[3]{n}}{sqrt n}right)=sqrt[3]{n}+frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$
therefore
$$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$
edited Nov 22 at 23:45
amWhy
191k28224439
answered Nov 22 at 19:59
gimusi
92.8k94495
May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
– William McGonagall
Nov 27 at 7:08
1
@WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
– gimusi
Nov 27 at 7:14
Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
– William McGonagall
Nov 27 at 7:26
@WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
– gimusi
Nov 27 at 7:37
Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
– William McGonagall
Nov 27 at 7:42
add a comment |
up vote
3
down vote
By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have
$$sqrt[3]{n+sqrt{n}}=sqrt[3]{n}, left(1+frac1{sqrt n}right)^frac13=sqrt[3]{n}+frac{sqrt[3]{n}}{3sqrt n}+oleft(frac{sqrt[3]{n}}{sqrt n}right)=sqrt[3]{n}+frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$
therefore
$$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$
edited Nov 22 at 23:45
amWhy
191k28224439
answered Nov 22 at 19:59
gimusi
92.8k94495
May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
– William McGonagall
Nov 27 at 7:08
1
@WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
– gimusi
Nov 27 at 7:14
Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
– William McGonagall
Nov 27 at 7:26
@WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
– gimusi
Nov 27 at 7:37
Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
– William McGonagall
Nov 27 at 7:42
add a comment |
up vote
3
down vote
up vote
3
down vote
By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have
$$sqrt[3]{n+sqrt{n}}=sqrt[3]{n}, left(1+frac1{sqrt n}right)^frac13=sqrt[3]{n}+frac{sqrt[3]{n}}{3sqrt n}+oleft(frac{sqrt[3]{n}}{sqrt n}right)=sqrt[3]{n}+frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$
therefore
$$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$
edited Nov 22 at 23:45
amWhy
191k28224439
answered Nov 22 at 19:59
gimusi
92.8k94495
By first order binomial expansion $(1+x)^r=1+rx + o(x)$, we have
$$sqrt[3]{n+sqrt{n}}=sqrt[3]{n}, left(1+frac1{sqrt n}right)^frac13=sqrt[3]{n}+frac{sqrt[3]{n}}{3sqrt n}+oleft(frac{sqrt[3]{n}}{sqrt n}right)=sqrt[3]{n}+frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$
therefore
$$sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{1}{3n^frac16}+oleft(frac{1}{n^frac16}right)$$
edited Nov 22 at 23:45
amWhy
191k28224439
answered Nov 22 at 19:59
gimusi
92.8k94495
edited Nov 22 at 23:45
amWhy
191k28224439
edited Nov 22 at 23:45
amWhy
191k28224439
edited Nov 22 at 23:45
amWhy
191k28224439
191k28224439
answered Nov 22 at 19:59
gimusi
92.8k94495
answered Nov 22 at 19:59
gimusi
92.8k94495
answered Nov 22 at 19:59
gimusi
92.8k94495
92.8k94495
May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
– William McGonagall
Nov 27 at 7:08
1
@WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
– gimusi
Nov 27 at 7:14
Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
– William McGonagall
Nov 27 at 7:26
@WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
– gimusi
Nov 27 at 7:37
Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
– William McGonagall
Nov 27 at 7:42
add a comment |
May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
– William McGonagall
Nov 27 at 7:08
1
@WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
– gimusi
Nov 27 at 7:14
Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
– William McGonagall
Nov 27 at 7:26
@WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
– gimusi
Nov 27 at 7:37
Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
– William McGonagall
Nov 27 at 7:42
May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
– William McGonagall
Nov 27 at 7:08
May I ask why $(1+x)^r=1+rx+o(x)$ when $r$ is fractional?
– William McGonagall
Nov 27 at 7:08
1
1
@WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
– gimusi
Nov 27 at 7:14
@WilliamMcGonagall Yes of course! By binomial theorem $(1+x)^r=1+rx+r(r-1)^x^2+...=1+rx+o(x). Refer also to Binomial Series
– gimusi
Nov 27 at 7:14
Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
– William McGonagall
Nov 27 at 7:26
Thanks for the pointer. The binomial theorem for fractional exponent looks mysterious to me.
– William McGonagall
Nov 27 at 7:26
@WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
– gimusi
Nov 27 at 7:37
@WilliamMcGonagall It can be derived by Taylor's expansion, are you aware about it?
– gimusi
Nov 27 at 7:37
Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
– William McGonagall
Nov 27 at 7:42
Yes, but I wasn't familiar with the order of growth of $binom{r}{k}$ and so I didn't know what the radius of convergence of the series would be.
– William McGonagall
Nov 27 at 7:42
add a comment |
up vote
3
down vote
Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$
We have that
$$0< sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{sqrt[3]{n}}{sqrt{n}}cdotfrac{
sqrt[3]{1+frac{1}{sqrt{n}}}-1}{frac{1}{sqrt{n}}}leq frac{1}{sqrt[6]{n}}cdot frac{1}{3}.$$
where we used the Bernoulli's inequality
$$(1+x)^rleq 1+rx$$
with $r=1/3in(0,1)$ and $x=1/sqrt{n}$.
Can you take it from here?
P.S. By using this approach we are also able to find
$$limlimits_{ntoinfty}left((n+sqrt{n})^r-n^rright)$$
for any $r<1/2$. Note that if we replace $sqrt[3]{ }$ with say $sqrt[5]{ }$ then the "algebraic way" could be very annoying!
answered Nov 22 at 19:30
Robert Z
92.4k1058129
That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
– SacredScout
Nov 22 at 19:37
1
Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
– Robert Z
Nov 22 at 19:39
I must have been blind..... Got that! I still have to admit that I don't get the denominator.
– SacredScout
Nov 22 at 19:45
$sqrt{n}cdot frac{1}{sqrt{n}}=1$...
– Robert Z
Nov 22 at 19:46
It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
– SacredScout
Nov 27 at 15:23
add a comment |
up vote
3
down vote
Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$
We have that
$$0< sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{sqrt[3]{n}}{sqrt{n}}cdotfrac{
sqrt[3]{1+frac{1}{sqrt{n}}}-1}{frac{1}{sqrt{n}}}leq frac{1}{sqrt[6]{n}}cdot frac{1}{3}.$$
where we used the Bernoulli's inequality
$$(1+x)^rleq 1+rx$$
with $r=1/3in(0,1)$ and $x=1/sqrt{n}$.
Can you take it from here?
P.S. By using this approach we are also able to find
$$limlimits_{ntoinfty}left((n+sqrt{n})^r-n^rright)$$
for any $r<1/2$. Note that if we replace $sqrt[3]{ }$ with say $sqrt[5]{ }$ then the "algebraic way" could be very annoying!
answered Nov 22 at 19:30
Robert Z
92.4k1058129
That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
– SacredScout
Nov 22 at 19:37
1
Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
– Robert Z
Nov 22 at 19:39
I must have been blind..... Got that! I still have to admit that I don't get the denominator.
– SacredScout
Nov 22 at 19:45
$sqrt{n}cdot frac{1}{sqrt{n}}=1$...
– Robert Z
Nov 22 at 19:46
It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
– SacredScout
Nov 27 at 15:23
add a comment |
up vote
3
down vote
up vote
3
down vote
Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$
We have that
$$0< sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{sqrt[3]{n}}{sqrt{n}}cdotfrac{
sqrt[3]{1+frac{1}{sqrt{n}}}-1}{frac{1}{sqrt{n}}}leq frac{1}{sqrt[6]{n}}cdot frac{1}{3}.$$
where we used the Bernoulli's inequality
$$(1+x)^rleq 1+rx$$
with $r=1/3in(0,1)$ and $x=1/sqrt{n}$.
Can you take it from here?
P.S. By using this approach we are also able to find
$$limlimits_{ntoinfty}left((n+sqrt{n})^r-n^rright)$$
for any $r<1/2$. Note that if we replace $sqrt[3]{ }$ with say $sqrt[5]{ }$ then the "algebraic way" could be very annoying!
answered Nov 22 at 19:30
Robert Z
92.4k1058129
Alternative approach where we do not use of the identity $$a^3-b^3=(a-b)(a^2+ab+b^2).$$
We have that
$$0< sqrt[3]{n+sqrt{n}}-sqrt[3]{n}=frac{sqrt[3]{n}}{sqrt{n}}cdotfrac{
sqrt[3]{1+frac{1}{sqrt{n}}}-1}{frac{1}{sqrt{n}}}leq frac{1}{sqrt[6]{n}}cdot frac{1}{3}.$$
where we used the Bernoulli's inequality
$$(1+x)^rleq 1+rx$$
with $r=1/3in(0,1)$ and $x=1/sqrt{n}$.
Can you take it from here?
P.S. By using this approach we are also able to find
$$limlimits_{ntoinfty}left((n+sqrt{n})^r-n^rright)$$
for any $r<1/2$. Note that if we replace $sqrt[3]{ }$ with say $sqrt[5]{ }$ then the "algebraic way" could be very annoying!
answered Nov 22 at 19:30
Robert Z
92.4k1058129
edited Nov 27 at 6:34
answered Nov 22 at 19:30
Robert Z
92.4k1058129
answered Nov 22 at 19:30
Robert Z
92.4k1058129
answered Nov 22 at 19:30
Robert Z
92.4k1058129
92.4k1058129
That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
– SacredScout
Nov 22 at 19:37
1
Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
– Robert Z
Nov 22 at 19:39
I must have been blind..... Got that! I still have to admit that I don't get the denominator.
– SacredScout
Nov 22 at 19:45
$sqrt{n}cdot frac{1}{sqrt{n}}=1$...
– Robert Z
Nov 22 at 19:46
It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
– SacredScout
Nov 27 at 15:23
add a comment |
That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
– SacredScout
Nov 22 at 19:37
1
Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
– Robert Z
Nov 22 at 19:39
I must have been blind..... Got that! I still have to admit that I don't get the denominator.
– SacredScout
Nov 22 at 19:45
$sqrt{n}cdot frac{1}{sqrt{n}}=1$...
– Robert Z
Nov 22 at 19:46
It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
– SacredScout
Nov 27 at 15:23
That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
– SacredScout
Nov 22 at 19:37
That looks really plausible and I could basically conclude that it converges to 0 with your term. But I have no Idea how I could rewrite the term in your manner, so that would sadly not help me in other tasks :(.... How did you rewrite it like that, by using which rules?
– SacredScout
Nov 22 at 19:37
1
1
Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
– Robert Z
Nov 22 at 19:39
Do you agree that $sqrt[3]{n}cdot ( sqrt[3]{1+frac{1}{sqrt{n}}}-1)=sqrt[3]{n+sqrt{n}}-sqrt[3]{n}$? Recall that $sqrt[3]{a}sqrt[3]{b}=sqrt[3]{ab}$.
– Robert Z
Nov 22 at 19:39
I must have been blind..... Got that! I still have to admit that I don't get the denominator.
– SacredScout
Nov 22 at 19:45
I must have been blind..... Got that! I still have to admit that I don't get the denominator.
– SacredScout
Nov 22 at 19:45
$sqrt{n}cdot frac{1}{sqrt{n}}=1$...
– Robert Z
Nov 22 at 19:46
$sqrt{n}cdot frac{1}{sqrt{n}}=1$...
– Robert Z
Nov 22 at 19:46
It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
– SacredScout
Nov 27 at 15:23
It is much better now, but still really tough for me to understand. Would it be another questions I would have great problems with it, no doubt. I started studying 2 months ago, so eventually I will fully understand that as well. Thanks, Sir!
– SacredScout
Nov 27 at 15:23
add a comment |
up vote
2
down vote
As you suggested,
begin{align}
sqrt[3]{n+sqrt n} - sqrt[3]{n} &= frac{n+sqrt n - n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}\
&= frac{sqrt n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}.
end{align}
The fastest growing term is evidently $n^{2/3}$. Can you finish?
answered Nov 22 at 19:31
MisterRiemann
5,7131624
add a comment |
up vote
2
down vote
As you suggested,
begin{align}
sqrt[3]{n+sqrt n} - sqrt[3]{n} &= frac{n+sqrt n - n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}\
&= frac{sqrt n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}.
end{align}
The fastest growing term is evidently $n^{2/3}$. Can you finish?
answered Nov 22 at 19:31
MisterRiemann
5,7131624
add a comment |
up vote
2
down vote
up vote
2
down vote
As you suggested,
begin{align}
sqrt[3]{n+sqrt n} - sqrt[3]{n} &= frac{n+sqrt n - n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}\
&= frac{sqrt n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}.
end{align}
The fastest growing term is evidently $n^{2/3}$. Can you finish?
answered Nov 22 at 19:31
MisterRiemann
5,7131624
As you suggested,
begin{align}
sqrt[3]{n+sqrt n} - sqrt[3]{n} &= frac{n+sqrt n - n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}\
&= frac{sqrt n}{(n+sqrt n)^{2/3}+sqrt[3]{n(n+sqrt n)} + n^{2/3}}.
end{align}
The fastest growing term is evidently $n^{2/3}$. Can you finish?
answered Nov 22 at 19:31
MisterRiemann
5,7131624
answered Nov 22 at 19:31
MisterRiemann
5,7131624
answered Nov 22 at 19:31
MisterRiemann
5,7131624
answered Nov 22 at 19:31
MisterRiemann
5,7131624
5,7131624
add a comment |
add a comment |
up vote
0
down vote
Consider the function $f(x)=x^{1/3}$. By the mean value theorem there's a number $yin (n, n+sqrt n)$ such that
$$
f(n+sqrt n) - f(n) = f'(y)(n+sqrt n - n)= frac{y^{-2/3}}{3}sqrt n<n^{-2/3}sqrt n=n^{-1/6}to 0.
$$
answered Nov 24 at 8:21
User
483514
add a comment |
up vote
0
down vote
Consider the function $f(x)=x^{1/3}$. By the mean value theorem there's a number $yin (n, n+sqrt n)$ such that
$$
f(n+sqrt n) - f(n) = f'(y)(n+sqrt n - n)= frac{y^{-2/3}}{3}sqrt n<n^{-2/3}sqrt n=n^{-1/6}to 0.
$$
answered Nov 24 at 8:21
User
483514
add a comment |
up vote
0
down vote
up vote
0
down vote
Consider the function $f(x)=x^{1/3}$. By the mean value theorem there's a number $yin (n, n+sqrt n)$ such that
$$
f(n+sqrt n) - f(n) = f'(y)(n+sqrt n - n)= frac{y^{-2/3}}{3}sqrt n<n^{-2/3}sqrt n=n^{-1/6}to 0.
$$
answered Nov 24 at 8:21
User
483514
Consider the function $f(x)=x^{1/3}$. By the mean value theorem there's a number $yin (n, n+sqrt n)$ such that
$$
f(n+sqrt n) - f(n) = f'(y)(n+sqrt n - n)= frac{y^{-2/3}}{3}sqrt n<n^{-2/3}sqrt n=n^{-1/6}to 0.
$$
answered Nov 24 at 8:21
User
483514
answered Nov 24 at 8:21
User
483514
answered Nov 24 at 8:21
User
483514
answered Nov 24 at 8:21
User
483514
483514
add a comment |
add a comment |
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