Why does this sum converge? I know that the following sum converges to 2 via WolframAlpha, but I am not sure why.  
$$\sum_{k=1}^\infty k \left[\frac{2}{k} - \frac{4}{k+1} + \frac{2}{k+2}\right] = 2$$
WolframAlpha gives the following partial sum formula:
$$\sum_{k=1}^n k \left[\frac{2}{k} - \frac{4}{k+1} + \frac{2}{k+2}\right] = \frac{2n}{n+2}$$
I would intuitively guess that the result of the partial sum formula is 2 for $n = \infty$. Where did that partial sum formula come from? Can someone help me build some intuition here?
 A: $$\begin{align}k\left(\frac 2k-\frac{4}{k+1}+\frac{2}{k+2}\right)&=2-\frac{4k}{k+1}+\frac{2k}{k+2}\\&=2-4\left(1-\frac{1}{k+1}\right)+2\left(1-\frac{2}{k+2}\right)\\&=4\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\end{align}$$
A: Let the sum of series be denoted by $S$, the term corresponding to each k be $s_k$. Thus each $s_k$ is made of three terms.
$S=\sum s_k=\sum k(\frac{2}{k}-\frac{4}{k+1}+\frac{2}{k+2})$
$=\sum\limits_{k=1}^{\infty} (2-\frac{4k}{k+1}+\frac{2k}{k+2})$
Notice that last term of each $s_k$ and first 2 terms of $s_{k+1}$ have a sum $0$ as seen below:  $\frac{2k}{k+2}+2-\frac{4(k+1)}{k+2}=2 +\frac{-2k-4}{k+2}=0$ 
Taking first two terms of $s_0$ and last term of $s_n$ as $n$ goes to infinity:
$S=\lim_{n \to \infty} (2-\frac{4*1}{1+1}+\frac{2n}{n+2})=\lim_{n \to \infty} \frac{2n}{n+2}=2$ 
A: You can start by doing the things you can do. They don't always help, but doing something is better than doing nothing, and you really should try out all of your leads before you declare yourself stuck on a problem.
One obvious thing that you can do is to simply simplify the terms; one such simplification is to combine the fractions. And it turns out that
$$ k \left[\frac{2}{k} - \frac{4}{k+1} + \frac{2}{k+2}\right] = \frac{4}{(k+1)(k+2)} $$
from which the problem is nearly trivial if you're at all familiar with similar examples.
