Question: I found this series in my calculus book: $$\sum\_{n=1}^{\infty}(-1)^{n+1} \frac{1}{5^nn}$$
The directions are in the title of this question, but I can't think of any functions whose power series looks anything like that when evaluated at a point. Hints are appreciated, because I like to work these out on my own.

Answer: 
$$\log(1-x)=-\sum\_{n\geqslant1}\frac{x^n}n$$
From Did's solution, we have

$$\log(1+x)=-\sum\_{n\geqslant1}\frac{(-x)^n}n=\sum\_{n\geqslant1}(-1)^{n+1}\frac{x^n}n$$

Setting $x=1/5$, we obtain:

$$\log(6/5)=\sum\_{n\geqslant1}(-1)^{n+1}\frac{1}{5^n n}$$