TITLE: Convergence or divergence of the series $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...$
QUESTION [2 upvotes]: I'm taking undergrad real analysis. I'm trying to come up with a general formula for the series
\begin{equation}
1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...
\end{equation} and cannot understand how to get the negative term to only show up with powers of $3$. Or is there another way to test the series. any help would be appreciated.

REPLY [0 votes]: $$ \begin{align} 
&\quad\, \sum_{n=0}^{\infty}\left[+\frac{1}{3n+1}+\frac{1}{3n+2}\color{red}{-}\frac{1}{3n+3}\right] \\[2mm] 
&= \sum_{n=0}^{\infty}\left[\frac{1}{3n+1}+\frac{1}{3n+2}\color{red}{-}\frac{2}{3n+3}+\frac{1}{3n+3}\right] \\[2mm] 
&= \sum_{n=0}^{\infty}\left[\frac{1}{3n+1}+\frac{1}{3n+2}\color{red}{-}\frac{2}{3n+3}\right]+\sum_{n=0}^{\infty}\left[\frac{1}{3n+3}\right] \\[2mm] 
&= \log{3}+\frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{n} \color{red}{\,\,\rightarrow\,\,\infty} \\[2mm] 
\end{align} $$
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