The series cannot converge absolutely, because the harmonic series diverges.

We will show that the series converges. The expression \(\frac{n(n+1)}2\) is odd when \(n\) gives a remainder of \(1\) or \(2\) when divided by four, and even in other cases. So the series begins with two negative terms, followed by two positive terms, and then again two negative terms etc. The idea behind our solution is to add pairs of consecutive terms with the same sign so that we can use Leibniz's test.

For odd \(n\), i.e. \(n = 2k - 1\) let us look at the \(n\)-th and \((n+1)\)-th. terms. These are \(\displaystyle \frac{(-1)^{\frac{(2k-1)2k}2}}{2k-1} \hbox{ and } \frac{(-1)^{\frac{2k(2k+1)}2}}{2k}.\)

As we have already determined, these two terms have the same sign. Concretely \((-1)^k\), as \((-1)^{2k-1} = (-1)^{2k+1} = -1\).

If we add these two terms we obtain \(\displaystyle (-1)^k\left(\frac{1}{2k-1} + \frac{1}{2k}\right)= (-1)^k\frac{4k - 1}{2k(2k-1)}. \)

The series \(\displaystyle \sum\limits_{k=1}^{\infty}(-1)^k\frac{4k - 1}{2k(2k-1)}. \) converges by Leibniz's test (check that the corresponding sequence of absolute values is non-increasing and has limit zero).

From there, we must make one more step to deduce that the given series converges. Define \(s_t\) to be the partial sum \(\displaystyle s_t = \sum\limits_{n=1}^{t} \frac{(-1)^{\frac{n(n+1)}2}}{n}. \)

To show that the series converges we must verify that the sequence of partial sums \(\{s_t\}\) converges. We know that the series \(\displaystyle \sum\limits_{k=1}^{\infty}(-1)^k\frac{4k - 1}{2k(2k-1)}. \) converges. But from its definition it follows that for now we only know that the sequence of even terms of partial sums \(\{s_{2m}\}\) converges. The last step is to realize that \(\displaystyle \lim\limits_{n \to \infty} \frac{(-1)^{\frac{n(n+1)}2}}{n} = 0, \) so for sufficiently large \(n\) the odd terms are close to the even terms, and so the entire given series converges (carefully consider the definition of a limit).