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I want to expand and test this $\{(n^3+1)^{1/3} - n\}$ for convergence/divergence.

The edited version is: Test for convergence $\sum_{n=1}^{\infty}\{(n^3+1)^{1/3} - n\}$

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Yes, it converges as $n\to 2$. But presumably that isn't what you mean, so you need to provide more information, and explain what you've tried or thought about so far. – Zev Chonoles Jun 29 '12 at 8:58
It is term of an infinite series (n is a natural no.). Interestingly in one book it is given to be divergent and in another it is convergent. BUT none tells how? – rajendra bakre Jun 29 '12 at 9:04
So, you want to consider $n\to \infty$. You should add that to the question - the "edit" button for your question is just below the "limit" and "convergence" tags, on the lower left. Now, what have you tried on this problem so far, if anything? – Zev Chonoles Jun 29 '12 at 9:08
This question is a duplicate of this question and consequently should be closed. – Zev Chonoles Jun 29 '12 at 9:10
Well! I am editing the question. I think I got the answer from avatar BUT I still have a little bit of doubt about his answer. I am writing the doubt as a comment to his answer – rajendra bakre Jun 29 '12 at 9:30
up vote 0 down vote accepted

By direct inspection, for every pair of real numbers $A$ and $B$, $$ A^3 - B^3 = (A-B) (A^2+AB+B^2). $$ Choose now $A=\sqrt[3]{n^3+1}$ and $B=n$. Then $$ (n^3+1)^{1/3} - n = \frac{n^3+1-n^3}{(n^3+1)^{2/3} + n (n^3+1)^{1/3}+n^2} \sim \frac{1}{n^2} $$ as $n \to +\infty$. The limit is therefore zero.

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Check the denominator, the $A^2$ term – leonbloy Jun 29 '12 at 10:33
I agree with you Siminore! But your A^2 and A term appears to be wrong in index. There should be '2/3' and '1/3' index respectively – rajendra bakre Jun 29 '12 at 10:44
Is it better now? Thank you! – Siminore Jun 29 '12 at 10:46
It's wrong, it should be either $\sim 1/(3 n^2)$, or $O(1/n^2)$ – leonbloy Jun 29 '12 at 12:14
We will never agree on the use of symbols :-) I learned that $a_n \approx b_n$ means that $a_n / b_n \to 1$, and that $a_n \sim b_n$ that $a_n / b_n$ tends to a limit different than zero (and finite). It seems hard to find a set of symbols that everybody can understand. – Siminore Jun 29 '12 at 13:09

The first thing we note is that, as $n\to \infty$ the term $(n^3+1)^{1/3}$ is of order $n$, which is cancelled by the other term, hence we must ask about the next order approximation.

Now, $ (n^3+1)^{1/3} = n \left(1 + \frac{1}{n^3}\right)^{1/3}$ and $(1+x)^{1/3} = 1 + \frac{1}{3}x + O(x^2)$, hence

$$ (n^3+1)^{1/3} - n = \frac{1}{3}\frac{1}{n^2}+O(n^{-5})$$

Hence, the series converge.

(If you want an explicit bound, you can just use $ (1+x)^{1/3} < 1+x$ for $x>0$, so you can compare to the series $\sum 1/n^2$ which you're supposed to know is convergent)

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Let $y=1/n$, then $\lim_{n\to \infty}{(n^3+1)}^{1/3}-n=\lim_{y\to0^+}\frac{(1+y^3)^{1/3}-1}{y}$. Using L'Hopital's Rule, this limit evaluates to $0$.Hence, the expression converges. You can also see that as $n$ increases, the importance of $1$ in the expression ${(n^3+1)}^{1/3}-n$ decreases and $(n^3+1)^{1/3}$ approaches $n$.Hence, the expression converges to $0$(as verified by limit). For the series part as @siminore pointed out,this difference ${(n^3+1)}^{1/3}-n$ is of the order of $1/n^2$,therefore, the sum of these is of the order of this sum :$\sum_{1}^{\infty}1/n^2$ which is $= {\pi}^2/6$. Thus the series is bounded and hence converges .

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Thanks avatar. The books I have do not mention that a series converges if its 'n'th term tends to 0 (or any finite no.). The sequence does converge but does it prove the series converges? – rajendra bakre Jun 29 '12 at 9:35
A necessary condition for a series $\sum_n a_n$ to converge is that $\lim_{n \to +\infty} a_n=0$. But this condition is far from being also sufficient. – Siminore Jun 29 '12 at 9:40
A sequence converges if its nth term converges,but for convergence of series, that nth term must converge to $0$ and even this is a necessary condition but not sufficient. – Aang Jun 29 '12 at 9:41
In that case (condition not sufficient) the convergence of the 'n'th term does not serve our purpose. Am I right? How about comparing the series with another known series? – rajendra bakre Jun 29 '12 at 9:53
That's what i have done in my answer.I have edited it.Check it. – Aang Jun 29 '12 at 9:54

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