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How do i check convergence/divergence of this series whose nth term is :

$$(n^3 + 1 )^{1/3}-n$$

godonichia
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4 Answers4

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Let : $ U_n = (n^3 + 1)^{\frac{1}{3}} -n = n*[ (1+\frac{1}{n^3})^{\frac{1}{3}} -1]$

$ (1+u)^a = 1 +a*u + o(u) $ ; when $u \rightarrow 0$

That gives: $ U_n = n*[(1+\frac{1}{3*n^3} +o(\frac{1}{n^3})) -1] = \frac{1}{3*n^2} +o(\frac{1}{n^2}) $

$\sum \frac{1}{n^2}$ converges, hence $\sum U_n$ converges as well

mvggz
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  • Can I know why I just got a down vote? My answer is good, just as @Yiorgos S. Smyrlis's – mvggz Nov 05 '14 at 13:26
  • I did not downvote but I would think that the downvote arose because you provided a fully worked out solution to an obvious homework problem only moments after it was asked. Many folks here object to that and, I think, with good reason. Perhaps a hint would be more appropriate? One could also ask - if you found the question interesting enough to answer, then why didn't you upvote it? – Mark McClure Nov 05 '14 at 13:32
  • Agreed, but if you give a hint you might as well give the whole solution since there is only one step. At least now he knows how to write this properly, and that applies to all similar questions :). I don't always answer questions because I find them interesting, I also answer just to help out the OP. In this case I don't always up-vote, maybe I should change that. It's just that I've often answered like this, up-voted, and got complete indifference from the OP. That was a bit annoying – mvggz Nov 05 '14 at 13:37
  • I have downvoted in this situation but only when I was involved in actively discussing hints. So I guess I think a down vote is reasonable. When I do (rarely) down vote, however, I also comment on why if no one else has yet. I would be more annoyed that the down voter didn't do that. – Mark McClure Nov 05 '14 at 13:49
  • Ok, good to know :) . But I didn't mean to spoil the question and the OP did not say if he wanted insights or a fully detailed answer. Next time I'll ask him first ;) – mvggz Nov 05 '14 at 14:02
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By the Bernoulli inequality: $$ (1+n^3)^{1/3}-n = n\left(\left(1+\frac{1}{n^3}\right)^{1/3}-1\right) \leq n\cdot\frac{1}{3n^3}=\frac{1}{3n^2}$$ hence: $$ \sum_{n=1}^{+\infty}\left((1+n^3)^{1/3}-n\right)\leq\frac{\zeta(2)}{3}=\frac{\pi^2}{18}.$$

Jack D'Aurizio
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Using the identity $\,a^3-b^3=(a-b)(a^2+ab+b^2)$, and consequently $\,a-b=\frac{a^3-b^3}{a^2+ab+b^2}$, we obtain (for $a=\sqrt[3]{n^3+1},\,b=n$) $$ (n^3+1)^{1/3}-n=\frac{(n^3+1)-n^3}{(n^3+1)^{2/3}+n(n^3+1)^{1/3}+n^2}= \frac{1}{(n^3+1)^{2/3}+n(n^3+1)^{1/3}+n^2}, $$ and $$ 0<\frac{1}{(n^3+1)^{2/3}+n(n^3+1)^{1/3}+n^2}\le\frac{1}{3n^2}. $$ Hence, our series converges due to the Comparison Test.

Yiorgos S. Smyrlis
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Key hint:

$$\left(1+{1\over n^3}\right)\le\left(1+{1\over3n^3}\right)^3$$

(which you can verify by expanding the right hand side).

Added later: Since others have given complete answers (and one of them has been accepted), I'll flesh out the hint here. The inequality implies

$$(n^3+1)^{1/3}-n=n\left(\left(1+{1\over n^3}\right)^{1/3}-1\right)\le n\left(\left(1+{1\over3n^3}\right)-1\right)={1\over3n^2}$$

The Comparison Test can now be brought to bear.

Barry Cipra
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