real analysis - Is this:$sum_{n=1}^{infty}{(-1)}^{frac{n(n-1)}{2}}frac{1}{n}$ a convergent series?

Is there someone who can show me how do I evaluate this sum : $$\sum_{n=1}^{\infty}{(-1)}^{\frac{n(n-1)}{2}}\frac{1}{n}$$

Note : In wolfram alpha show this result and in the same time by ratio test it's not a convince way to judg that is convergent series

Thank you for any help

Answer

The parity of $\frac{n(n-1)}{2}$ is 4-periodic. Thus the sequence $(-1)^{\frac{n(n-1)}{2}}$ equals to:

$$1, \, -1, \, -1, \, 1, \, 1, \, -1, \, -1, \, 1, \, 1, \, -1, \, -1, \, 1 , \cdots$$
The original series' partial sum truncated at $N$ equals to
$$\sum_{k=0}^{K} \left( \frac{1}{4k+1} - \frac{1}{4k+2} - \frac{1}{4k+3} + \frac{1}{4k+4}\right) + \sum_{i=1}^{N - 4K - 4}\frac{(-1)^{\frac{i(i-1)}{2}}}{4K + 4 + i}$$
where $K = \lfloor \frac{N}{4}\rfloor - 1$.

Then by a discussion on the partial sum we can conclude that the series is convergent.