Convergent/divergent series Is the following series divergent/convergent? 
$$S=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}-...$$
I think it is divergent since
$$
\begin{align}
S&>1-\frac{1}{2}-\frac{1}{2}+4\cdot\frac{1}{6}-\frac{4}{7}+\frac{5}{15}-\frac{6}{16}+...\\
&=1/3-4/7+1/3-6/16+...=\sum_{n=1}^\infty 1/3-\alpha_n
\end{align}
$$
where $\alpha_n=4/7, 6/16, 8/22$ which tends to 0, so the series on the right hand side is divergent. Is this the right answer? Thanks 

After placing brackets around the terms with the same sign the series becomes
$$1+\sum_{n=2}^\infty(-1)^{n+1}(H_{n(n+1)/2}-H_{n(n-1)/2}),$$
where $H_n=1+\frac12+\frac13+\ldots+\frac1n$ is the $n$-th harmonic number. Now using well-known formula $$H_n=\log n+\gamma+O\left(\frac{1}{n}\right),$$
we get
$$a_n=H_{n(n+1)/2}-H_{n(n-1)/2}=\log\frac{n+1}{n-1}+O\left(\frac{1}{n^2}\right)=$$
$$=\log\left(1+\frac{2}{n-1}\right)+O\left(\frac{1}{n^2}\right)=\frac{2}{n}+O\left(\frac{1}{n^2}\right).$$
Therefore the main term of the series with braces is $$2\frac{(-1)^{n+1}}{n}+O\left(\frac{1}{n^2}\right),$$
which means that the series converges (by Leibniz's and comparison tests), and so does the original one.
