Convergence of Infinite Series

Question

How can I determine that the following series is convergent:

$$ \sum_{x = 1}^{\infty}\left(\,\sqrt[3]{\,{x^{3} + 1}\,}\, - x\right) $$

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.

Answer 1

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 $0$.

Answer 2

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.