TITLE: Does $(1+\frac12-\frac13) + (\frac14+\frac15-\frac16)+(\frac17+\frac18-\frac19)+\cdots$ converge?
QUESTION [6 upvotes]: Does the series $$S=\left(1+\frac{1}{2}-\frac{1}{3} \right) + \left(\frac{1}{4}+\frac{1}{5}-\frac{1}{6} \right)+\left(\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)+\cdots$$ converge?

Here's my attempt at a solution: $$S = \sum_{n=1}^{\infty}\frac{1}{n}-2\sum_{n=1}^{\infty}\frac{1}{3n}=\sum_{n=1}^{\infty}\frac{1}{3n}=\frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{n}$$
As we can "rewrite" this series as one third of the harmonical series (that diverges), we conclude the divergence of $S$. 
Is this right? Which other convergence tests could be used?

REPLY [5 votes]: The idea is correct, but you cannot write like that, because the series involved are both divergent and generally it is moot to do arithmetic operations of divergent series. 
Let $H_n = \sum_1^n 1/k$, i.e. the $n$-th partial sum of the harmonic series, then the 
$$
S_n = H_{3n} - \frac 23 H_n, 
$$
then use the asymptotic expression $H_n = \log n + \gamma + \varepsilon_n$ where $\varepsilon_n \to 0 [n \to \infty]$, we have
$$
S_n = \log (3n) -\frac 23\log n + \frac 13 \gamma + \varepsilon_{3n} - \frac 23 \varepsilon_n = \log(3n^{1/3}) + \frac 13 \gamma + \alpha_n \xrightarrow{n \to \infty} +\infty. 
$$