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

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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.

Some of the posted answers are using rearrangements or regrouping, but these are treacherous techniques to apply to series which diverge absolutely (as this one plainly does).

Leibniz’ Theorem as usually stated only applies to alternating series, but we can modify the proof of Leibniz’ Theorem to cover this particular series.

Leibniz Theorem: A series of the form $\sum(-1)^n a_n$ wherein the $a_n$ are all positive and the $a_n$ decrease to 0 converges.

Proof: Let $S_k$ denote the sum of the first k terms. Now look at two consecutive partial sums, $S_k$ and $S_{k+1}$ The next one, $S_{k+2}$, clearly falls between these (as the series is alternating and decreasing). In this way we see that all subsequent partial sums fall between any two consecutive partial sums. As the gap between consecutive partial sums goes to 0 we see that the partial sums, and therefore the series, must converge.

As this series is not alternating, we can’t invoke the theorem casually. But the logic of the proof works just fine: for the given series we just observe that all subsequent partial sums are trapped between partial sums of the form $S_{4k+1}$ and $S_{4k+3}$ and the gaps between partial sums a fixed distance apart clearly goes to $0$.

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.

I post this answer because Dirichlet’s test has not been mentioned in any of the previous answers. Let $a_n=(-1)^{n(n-1)/2}$ and $b_n=1/n$. The partial sums of $a_n$ are bounded and $b_n$ is decreasing and converging to $0$. Dirichlet’s test implies the series is convergent.

Let $A_N$ be the $N$th partial sum of $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}.$

Let $B_N$ be the $N$th partial sum of $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2n}.$

Let $C_N$ be the $N$th partial sum of $\sum_{n=1}^{\infty}\frac{(-1)^{n(n-1)/2}}{n}.$

Then $C_{2N} = A_N + B_N.$ By the alternating series test, both $A_N,B_N$ converge. Hence $C_{2N}$ converges to some limit $S$. Now $C_{2N+1}=C_{2N+1}-C_{2N}+C_{2N}$ and $C_{2N+1}-C_{2N}=(\pm 1)/(2N+1) \to 0.$ Thus $C_{2N+1}\to S.$
A sequence whose even terms converge to $S$ and whose odd terms converge to $S$ itself converges to $S.$ Thus $C_N$ converges, which is the desired result.