Prove or disprove the series converge conditionally

Question

I have the following series - $$ \sum_{n = 1}^\infty {\frac {(-1)^{k(n)}}{n}} $$ when $$k(n)=\begin{cases} 1 &; \quad n \ \text{=$3m$, for $m$ natural number}\\ 2 &; \quad n \ \text{otherwise}\ . \end{cases}$$

It doesn't converge absolutely because it's similar to the harmonic series , but does it converge conditionally or not ? i would like to get some detailed explanation. Thanks.

Answer 1

So: $$ 1 + \frac 1 2 - \frac 1 3 + \frac 1 4 +\frac 1 5 - \frac 1 6 + \cdots$$

If you group the terms three at a time, the second term is greater than the third, so if you denote the sum up to $N$ by $S_N$, you get

$$S_{3K+1} > 1 + \frac 1 4 + \cdots + \frac 1 {3K+1}$$

You should be able to show that this subsequence goes to $+\infty$ so the original series is not convergent.

Answer 2

Condensation is enough: $$\begin{eqnarray*}\sum_{n=1}^{3N+1}\frac{ 1-2\cdot\mathbb{1}_{3\mathbb{Z}}(n) }{n}&=& 1+\sum_{k=1}^{N}\left(\frac{1}{3k-1}-\frac{1}{3k}+\frac{1}{3k+1}\right)\\&=&1+\frac{H_N}{3}+\sum_{k=1}^{N}\frac{2}{(3k-1)(3k)(3k+1)}\end{eqnarray*}$$ where the last term is bounded by $\sum_{k=1}^{+\infty}\frac{2}{(3k-1)(3k)(3k+1)}=-1+\log(3)$, but the previous term grows like $\frac{1}{3}\,\log N$, that is unbounded as $N\to +\infty$.