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  

After $ \ n = 1 \ $ , the exponents of (-1) are binomial coefficients which are "double-alternating" between even and odd integers.  So the series looks like
$$ 1 \ - \frac{1}{2} \ - \ \frac{1}{3} \ + \ \frac{1}{4} \ + \ \frac{1}{5} \ - \ \frac{1}{6} \ - \ \frac{1}{7} \ + \ \ldots \ \ . $$
As lulu I think properly objected to my separation of terms originally, since the absolute series here is the harmonic series, the "Riemann derangement" theorem may have something to say against it. It is perhaps safer then to group the terms as
$$ 1 \ + \ \sum_{k=1}^{\infty} \ (-1)^k\left( \ \frac{1}{2k} \ + \ \frac{1}{2k+1} \ \right) \ = \ 1 \ + \ \sum_{k=1}^{\infty} \ (-1)^k\left[ \ \frac{4k + 1}{2k \ (2k+1)} \ \right]  $$
$$ = \ 1 \ + \ \sum_{k=1}^{\infty} \ (-1)^k\left[ \ \frac{4k \ + \ 1}{4k^2 \ + \ 2k} \ \right] \ \ .  $$
The term $ \ b_k \ = \ \frac{4k \ + \ 1}{4k^2 \ + \ 2k} \ $ proves to be monotonically decreasing toward zero for $ \ k \ \ge \ 1 \ $ , so the series converges by the alternating-series test.  (And now my post looks a lot more like corindo's ...)
EDIT: Having thought about this a bit more, it occurred to me that I could have gathered this up a bit more (and now it comes still closer to what corindo is describing).  The alternating series could have simply been written as
$$ 1 \ + \ \sum_{k=1}^{\infty} \  \left( \ \frac{-1}{4k-2} \ + \ \frac{-1}{4k-1} \ + \ \frac{1}{4k} \ + \ \frac{1}{4k+1} \ \right)   $$
$$  = \ 1 \ - \ \sum_{k=1}^{\infty} \ \left[ \ \frac{32k^2 \ - \ 8k \ - \ 1}{4 \ x \ (2x-1) \ (4x-1) \ (4k+1)} \ \right] $$
$$ = \ 1 \ - \ \frac{1}{4} \ \sum_{k=1}^{\infty} \ \left( \ \frac{32k^2 \ - \ 8k \ - \ 1}{32k^4 \ - \ 16k^3 \ - \ 2k^2 \ + \ k} \ \right) \ \   $$
(with a little help from WolframAlpha).  This removes the "alternating-sign" behavior and allow us to apply the "limit comparison test" against $ \ \sum_{k=1}^{\infty} \ \frac{1}{k^2} \ $ . 
We can use this to set bounds on the sum (I've only pursued this a little at this point).  It is not hard to show that $ \ \frac{32k^2 \ - \ 8k \ - \ 1}{32k^4 \ - \ 16k^3 \ - \ 2k^2 \ + \ k} \ < \ \frac{2}{k^2} \ $ , from which we obtain
$$  1 \ - \ \frac{1}{4} \ \sum_{k=1}^{\infty} \ \left( \ \frac{32k^2 \ - \ 8k \ - \ 1}{32k^4 \ - \ 16k^3 \ - \ 2k^2 \ + \ k} \ \right) \  > \ 1 \ - \ \frac{1}{4} \ \sum_{k=1}^{\infty} \   \ \frac{2}{ k^2  }   $$
$$ = \ 1 \ - \ \frac{1}{2} \ \zeta(2) \ = \ 1 \ - \ \frac{\pi^2}{12} \ \ . $$
Others here are likely better able to set tighter bounds on the sum than I am at the moment. 
