How to compute the series $\sum_{n=0}^{\infty}{\frac{1}{(n+2)7^n}}$ I have to compute the series
$$\sum_{n=0}^{\infty}{\frac{1}{(n+2)7^n}}$$
It's easy to prove that the series converges using comparison test but how to compute it?

Consider
$$ S(x) = \sum_{n=0}^{\infty} \frac{x^n}{n+2}. $$
Your sum is $S(1/7)$. Then
$$ x S'(x)+2 S(x) = \sum_{n=0}^{\infty} \frac{(n+2)x^{n}}{n+2} = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} $$
An integrating factor for this differential equation is $x$, so we find that
$$ (x^2S)' = \frac{x}{1-x} = \frac{1}{1-x}-1 \\
x^2 S = C-x-\log{(1-x)}, $$
and setting $x=0$ implies that $C=0$. Hence
$$ S(x) = \frac{-1}{x^2}(x+\log{(1-x)}), $$
so
$$ S(1/7) = -49\left( \frac{1}{7} + \log{\left( \frac{6}{7} \right)} \right) $$
