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How can I determine that the following series is convergent:

$$ \sum_{x=1}^\infty \sqrt[3]{x^3+1}-x $$

I used the limit divergence test and I found that the limit of the nth term is zero. So that was of no use. Integral test is not worth as integration of this function is very difficult. Ratio Test and Root test are not appropriate either. Any help would be appreciated.

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Did you try Weierstrass M - test ? – Epsilon Apr 1 '13 at 12:50
up vote 3 down vote accepted

You can do that as Ron Gordon says, which was also my first hint. So here is a slightly less low tech approach using Taylor expansions. $$ \sqrt[3]{n^3+1}-n=n\left(\sqrt[3]{1+\frac{1}{n^3}}-1\right)=n\left( 1+\frac{1}{3}\frac{1}{n^3}+O\left(\frac{1}{n^6}\right) -1\right) $$ $$=\frac{1}{3}\frac{1}{n^2}+O\left(\frac{1}{n^5}\right)\sim \frac{1}{3n^2}. $$

So the series converges by limit comparison with $\sum_{n\geq 1}\frac{1}{n^2}$.

Note: I have simply used the Taylor expansion $(1+u)^\alpha=1+\alpha u+O(u^2)$ as $u$ tends to $0$.

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Nice way of doing it :) – Eddy Apr 1 '13 at 13:03

Use the relation

$$a^3-b^3=(a-b)(a^2+a b+b^2)$$

with $a=(x^3+1)^{1/3}$ and $b=x$.

The summand is equal to

$$\frac{1}{(x^3+1)^{2/3} + x (x^3+1)^{1/3} + x^{2}} \sim\frac{1}{3 x^2}$$

as $x \rightarrow \infty$. Then use the comparison test.

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I don't understand how you got the summand? – Eddy Apr 1 '13 at 13:01
Use the formula in the form $$a-b=\frac{a^3-b^3}{a^2+a b+b^2}$$ You should see that $a^3-b^3=1$. The denominator is what I have. – Ron Gordon Apr 1 '13 at 13:03
You need a $3$ in your last denominator, I think. – 1015 Apr 1 '13 at 13:04
Thanks for your method as well :) – Eddy Apr 1 '13 at 13:13

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