$$\sum\limits_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{H_n}$$
Where $H_n$ denotes the nth harmonic number.
I am a little confused because WolframAlpha says here that it is approximately $-0.626332$ but when I type it in slightly differently, it gives $-0.550242…$ I tried putting these numbers in the inverse symbolic calculator to find possible closed forms but I didn't get anything that seemed useful. I may just be overlooking something.
Also, how would I go about solving this by hand? I know it is equivalent to $\sum\limits_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{\sum\limits_{m=1}^{n}\frac{1}{m}}$ but I can't tell if that actually helps with anything. I've played around with it in Desmos a little and can't seem to figure anything out.
Best Answer
Using the Euler–Boole summation formula, $$ \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}{{H_n }}} \sim \sum\limits_{n = 1}^{N - 1} {\frac{{( - 1)^n }}{{H_n }}} + ( - 1)^N \frac{1}{2}\sum\limits_{k = 0}^\infty {\frac{{E_{2k} }}{{4^k (2k)!}}\left[ {\frac{{{\rm d}^{2k} }}{{{\rm d}x^{2k} }}\frac{1}{{\psi (x + 1/2) + \gamma }}} \right]_{x = N} } $$ as $N\to +\infty$. Here, $E_k$ denotes the Euler numbers, $\gamma$ is the Euler–Mascheroni constant, and $\psi(x)$ is the digamma function. Taking $N=100$ and keeping the first $5$ terms of the asymptotic expansion, I obtained $-0.626332482737912\ldots$ for the value of your sum.