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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.

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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 $S\approx 0.527273$.

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  • $\begingroup$ Thank you, great answer! Is it possible to prove the (ir)rationality of $ S $? $\endgroup$ – Vincenzo Oliva Oct 22 '14 at 9:42
  • $\begingroup$ Not necessarily asking you to do it, since it looks at least very hard, I'd like some advice on how doing it if it is possible. $\endgroup$ – Vincenzo Oliva Oct 22 '14 at 13:43

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