Series is convergent but it seems it is divergent? I have a series:
$$
\sum^\infty_{n=1}{\bigg(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+2}}\bigg)}
$$
and I thought it is a divergent series since
$$
\sum{\big(f(x)-g(x)\big)} = \sum{f(x)} - \sum{g(x)}
$$
and so the series equals to
$$
=\sum^\infty_{n=1}{\frac{1}{\sqrt{n}}}-\sum^\infty_{n=1}{\frac{1}{\sqrt{n+2}}}
$$
and we know that $\sum^\infty_{n=1}{\frac{1}{\sqrt{n}}}$ is divergent, so the whole series is divergent. But it turns out that it is convergent and the answer is $1+\frac{1}{\sqrt{2}}$.
How do you prove that it is convergent and calculate the answer? The only way I know how to compute an answer for a series is via a geometric series and there seems to be no way to make it into a geometric series!
 A: The rule you quoted above $\sum (f(x)-g(x)) = \sum f(x) - \sum g(x)$ is not correct if one or both of the series on the right is divergent.  In this case, both $\sum \frac{1}{\sqrt{n}}$ and $\sum \frac{1}{\sqrt{n+2}}$ are divergent.
As indicated in the comments, to evaluate this series, you should telescope the series.
A: Hint
Write
$$\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+2}}=\bigg(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\bigg)+\bigg(\frac{1}{\sqrt{n+1}}-\frac{1}{\sqrt{n+2}}\bigg)$$
and then telescope.
A: First to address your concern,
$$\sum_{n=1}^\infty \bigg(f(x)-g(x)\bigg)=\sum_{n=1}^\infty f(x)-\sum_{n=1}^\infty g(x)$$ 
is valid only when both $f$ and $g$ are convergent. An infinite sum is nothing more than the limit of partial sums, and the above is equivalent to writing,
$$\lim_{n\to\infty} (S_{n_1}-S_{n_2}) = \lim_{n\to\infty} S_{n_1} - \lim_{n\to\infty}S_{n_2}$$
which is only right when both the individual limits exist.

Now moving on to the problem at hand,

$$\sum_{n=1}^\infty \frac1{\sqrt{n}} -\frac1{\sqrt{n+2}}$$
$$S_n=1-\frac1{\sqrt{3}}+\frac1{\sqrt{2}}-\frac1{\sqrt{4}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\cdots +\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+2}}$$
$$S_n=1+\frac1{\sqrt{2}}-\frac{1}{\sqrt{n+2}}$$
Clearly, we have,
$$\lim_{n\to\infty} S_n=1+\frac{1}{\sqrt{2}}$$
A: Note:
$$\sum_{n = 1}^\infty \left[\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 2}}\right] = 1 - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{4}} - \frac{1}{\sqrt{6}} + \cdots = 1 + \frac{1}{\sqrt{2}}.$$
By the way, if $\sum f(n)$ and $\sum g(n)$ diverge, it does not mean $\sum [f(n) - g(n)]$ diverges because, for example,
$$\sum_{n = 1}^\infty \left[\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 2}}\right]$$
converges.
