Does $(1+\frac12-\frac13) + (\frac14+\frac15-\frac16)+(\frac17+\frac18-\frac19)+\cdots$ converge?

Question

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?

Answer 1

Your answer is correct, but your reasoning is not. Order matters. You should write the sum as follows: $$\sum_{n=0}^\infty\left(\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{1}{3n+3}\right)$$ Simplify what is in the parentheses and then evaluate in the usual way.

Oh, and it is true that order does not matter if all the terms are positive. But when some terms are positive and others negative, you have to be more careful.

Answer 2

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 $$ S_n = \log (3n) -\frac 23\log n + \frac 13 \gamma + \varepsilon_{3n} - \frac 23 \varepsilon_n = \log(3n^{1/3}) + \frac 13 \gamma + \alpha_n \xrightarrow{n \to \infty} +\infty. $$