Tell if a sum is convergent $\sum\limits_{n=1}^\infty \frac{2}{n(n+1)}$ $$\sum\limits_{n=1}^\infty \frac{2}{n(n+1)}$$
I tried to solve this by saying that $$\frac{2}{n(n+1)} = \frac{2}{n} - \frac{2}{n+1}$$
I then made two sums like this:
$$\sum\limits_{n=1}^\infty \frac{2}{n} - \sum\limits_{n=1}^\infty \frac{2}{n+1}$$
And since we know that $1/n$ is divergent I said that $\sum\limits_{n=1}^\infty \frac{2}{n(n+1)}$ would also be divergent, which for some reason was wrong. So my question is simply what am I doing wrong?
 A: Once you have split the fraction up, write down the first few terms of the initial sum in the split form. Can you find an explicit form for the sum to $N$? Then say what happens as $N\to \infty$?
What you have done is to split a sum with positive terms into one with both positive and negative terms. This second sum can only be safely rearranged as you have done if it is absolutely convergent. Your observation at the end shows that it isn't absolutely convergent, and demonstrates that the rearrangement was illegitimate.
A: This answer will consist of two parts: first, I will explain why your solution is wrong, then give you advice how to solve your problem correctly.

You say that because $a_n=b_n + c_n$ and the series
$$\sum_{n=1}^\infty b_n\\
\sum_{n=1}^\infty c_n$$
both diverge, then $$\sum_{n=1}^\infty a_n$$
must also diverge.
By that logic, take the series
$$a_n = \frac{1}{2^n}$$
a famous example of a converging series. You can then write $a_n = \left(\frac1{2^n}-1\right) + 1$, so $a_n=b_n + c_n$ if you define $b_n=\frac1{2^n} - 1$ and $c_n=1$.
It is clear that both $\sum b_n$ and $\sum c_n$ diverges, but still, $a_n$ can be summed. Therefore, your argument is not valid.

Now, how can we prove that the series converges? Well, you have a series of tests available: the ratio test, the root test, and the Raabe's test are probably known to you. I suggest you try them out (not all will work, but one should).
Another way of doing this is noticing that $\frac{2}{n(n+1)}$ is somewhat similar (asymptotically) to $\frac1{n^2}$. So, maybe, you can find some constant $C$ for which you could say that $$\frac{2}{n(n+1)}\leq C\cdot \frac1{n^2}?$$
Yety another thing you could do is write the sum (after writing $\frac{1}{n(n+1)}$ as $\frac1n -\frac1{n+1}$) down (say, the first $4$ sumands) to see what happens. You may be pleasantly surprised at the result.
A: Beside what has been said in comments and answers, you could have noticed that $$\sum\limits_{n=1}^\infty \frac{2}{(n+1)^2}<\sum\limits_{n=1}^\infty \frac{2}{n(n+1)}<\sum\limits_{n=1}^\infty \frac{2}{n^2}$$ Admitting you know the result of the summations of the extremes, you then have $$\frac{\pi ^2}{3}-2<\sum\limits_{n=1}^\infty \frac{2}{n(n+1)}<\frac{\pi ^2}{3}$$
A: There are multiple ways you can see that this series in fact is convergent.
You can correctly split up the fraction such that
$$\frac{2}{n(n+1)}=\frac{2}{n}-\frac{2}{n+1}. $$
However, you should here notice that almost all terms will cancel out. This is called a telescoping series. In fact, you will get
$$\sum\limits_{n=1}^N[a_n-a_{n-1}] = a_N-a_{0}.$$
You can also see that 
$$\frac{1}{n(n+1)}\sim\frac{1}{n^2} $$
and I'm guessing you know that the series of the latter is convergent. Here you can use the limit comparison test.
A: Here are the steps,
$$ \sum\limits_{n=1}^\infty \frac{2}{n(n+1)} $$
$$= 2\sum\limits_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right) $$
$$=2\lim\limits_{m\to\infty}\sum\limits_{n=1}^m \left(\frac{1}{n}-\frac{1}{n+1}\right) $$
$$=2\lim\limits_{m\to\infty} \left(1-\frac12+\frac12-\frac13 + \cdots +\frac{1}{m}-\frac{1}{m+1}\right) $$
$$=2\lim\limits_{m\to\infty} \left(1-\frac{1}{m+1}\right) $$
$$= 2(1-0) = 2$$
Therefore, this series converges to $2$.
A: divergent minus divergent may be convergent: sigma (1/n-1/n)=0 while sigma 1/n and sigma 1/n are divergent. The given series is a convergent telescoping series and it is a difference of two divergent series. 
A: Let's write the partial sum $\sum\limits_{n=1}^k \frac{2}{n} + \sum\limits_{n=1}^k \frac{-2}{n+1} = \frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}+...-\frac{2}{k}+\frac{2}{k}-\frac{2}{k+1}.\\ $
It remains only  $2-\frac{2}{k+1} \rightarrow 2  $ for $ k \rightarrow \infty $.
So by the definition of the convergence of a series it is convergent.
You cannot say that in a sum of two series if one diverges than the enterely sum diverges. You can only say, for example, in a sum of two series, if one converges and the other diverges the sum diverges, it is simple to demonstrate.
