<s>[INST] 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 
 [/INST] 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.
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