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!

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.
