Infinite series with addition and subtraction of fractions 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.
 A: HINT: Write out a few terms and see that it telescopes.
A: 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. 
A: Hint:
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.
A: If you split it into two infinite series.
$\sum_{n=1}^\infty \frac {2}{n^{1/3}}$ - $\sum_{n=1}^\infty \frac {2}{(n+1)^{1/3}}$
You can then see that they are the same just with the terms shifted over by 1.  By this, I mean that the nth+1 term of $\sum_{n=1}^\infty \frac {2}{n^{1/3}}$ is the same as the nth term of $\sum_{n=1}^\infty \frac {2}{(n+1)^{1/3}}$.  
With this understanding it's quite simple to take the difference of the two infinite series (all the terms cancel except for where n=1) 
So, the only term left is  $\sum_{n=1}^1 \frac {2}{n^{1/3}}$ = $\frac {2}{1^{1/3}}$ = 2.
