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The question is to find whether it converges or diverges. If it converges, find what it converges to?

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

I tried splitting up into two summations but that didn't help.

I think you can use the comparison test but if I use $\large\frac{2}{n^{1/3}}$ to compare, it is a $p$-series with $p < 1$, which diverges.

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up vote 4 down vote accepted

HINT: Write out a few terms and see that it telescopes.

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You have already been given two nice ways to solve the problem, so it is time for an ugly way.

Bring to a common denominator. Multiply top and bottom by $(n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3}$ and use the identity $x^3-y^3=(x-y)(x^2+xy+y^2)$ with $x=(n+1)^{1/3}$ and $y=n^{1/3}$. Our expression simplifies to $$\frac{2}{n^{1/3}(n+1)^{1/3}\left( (n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3}\right)}.$$ This is $\lt \dfrac{1}{n^{4/3}}$.

Another way: The following is better, no "magic" identities. Let $f(x)=\dfrac{2}{x^{1/3}}$. Then $f'(x)=-\dfrac{2}{3x^{4/3}}$. By the Mean Value Theorem, there is a $c$ between $n$ and $n+1$ such that $$f(n+1)-f(n)=-\frac{2}{3c^{4/3}}.$$ It follows that our $n$-th term, which is $f(n)-f(n+1)$, is $\lt \dfrac{2}{3n^{4/3}}$. We conclude convergence from the Comparison Test.

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Brian already gave a great hint, but there is yet another approach: set $b_{2n} = \frac{2}{n^{1/3}}$, $b_{2n+1} = -\frac{2}{(n+1)^{1/3}}$, so that $a_n = b_{2n} + b_{2n+1}$ and use Alternating Series Test.

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