# Prove $\sum_{n=1}^\infty \sum_{m=2^{n+1}} ^\infty \frac{(-1)^mn}{m}$ converges and evaluate it.

Noting that the series inside is a part of the alternating harmonic series multiplied by $-n$, we get $\displaystyle \sum_{m=2^{n+1}} ^\infty \frac{(-1)^mn}{m}=\sum_{m=1} ^\infty \frac{(-1)^mn}{m} -\sum_{m=1}^{2^{n+1}-1}\frac{(-1)^mn}{m}=\sum_{m=1}^{2^{n+1}-1}\frac{(-1)^{m-1}n}{m}-n\log2$, hence the initial sum is equivalent to$$\displaystyle \sum_{n=1}^\infty \sum_{m=1} ^{2^{n+1}-1} \frac{(-1)^{m-1}n}{m}-n\log2. \$$ It is obvious that the $n$-th term tends to $0$, so the sum can converge, and in fact I'm quite sure it does, having computed some partial sums as well. Finally, using the Lerch transcendent $\Phi$, we can rewrite it as $$\displaystyle \sum_{n=1}^\infty n \Phi(-1,1,2^{n+1}),$$ though this is perhaps nicer than useful.

Using $$\frac{1}{2k}-\frac{1}{2k+1} = \frac{1}{4k^2+2k}$$ write $M=m/2$ to get $$T(n) = \sum_{m=2^{n+1}}^\infty \frac{(-1)^m}{m} = \sum_{M=2^n}^\infty \frac{1}{4M^2+2M}$$ from which it follows that $$\int_{t=2^n}^\infty (2t+1)^{-2}~dt < \sum_{M=2^n}^\infty \frac{1}{(2M+1)^2} < T(n) < \sum_{M=2^n}^\infty \frac{1}{4M^2} < \int_{t=2^n-1}^\infty (2t)^{-2}~dt$$ and we get the bounds $$\frac{1}{2^{n+2}+2} < T(n) < \frac{1}{2^{n+2}-4}$$ It is thus clear that $$S = \sum_{n=1}^\infty \sum_{m=2^{n+1}} ^\infty \frac{(-1)^mn}{m} = \sum_{n=1}^\infty nT(n)$$ converges. Furthermore, we can use $$T(n) = \sum_{M=2^n}^{2^N-1} \frac{1}{4M^2+2M} + T(N)$$ with the bounds to get an error of order $O(2^{-N})$ and a numerical approximation $S\approx 0.527273$.
• Thank you, great answer! Is it possible to prove the (ir)rationality of $S$? – Vincenzo Oliva Oct 22 '14 at 9:42