In fact the series would converge even if$\ m$ were not natural, I just wanted to state that it is natural in my case. I have found the partial sum formula of$\ S_0$,$\displaystyle \sum_{n=1}^k \frac{n}{2^n} =\frac{2^{k+1}-k-2}{2^k}$, thus easily obtaining$\ S_0=2$. Then, since $\displaystyle \frac{1}{2^n+m}=\frac{1}{2^n}-\frac{m}{2^n\left(2^n+m\right)}$, I know $\displaystyle S_m= 2-\sum_{n=1}^\infty \frac{mn}{2^n\left(2^n+m\right)}$, though I'm not sure it is a convenient path to study the last series.

  • $\begingroup$ I don't see the last form helping, in particular the $m$ moves through the top. $\endgroup$ – djechlin Oct 18 '14 at 15:28
  • $\begingroup$ A quick Python script says that $S_m$ tends to $0$ as $m$ increases, but that is obvious. $\endgroup$ – Soham Chowdhury Oct 18 '14 at 15:28
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    $\begingroup$ There's a serious argument you can learn a lot more calculating 5 terms by hand than asking a computer to do infinity terms... $\endgroup$ – djechlin Oct 18 '14 at 15:35
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    $\begingroup$ Then it's a graph of your computing difficulties. $S_m$ is obviously decreasing. $\endgroup$ – djechlin Oct 18 '14 at 15:41
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    $\begingroup$ $S_1 = (\vartheta_3(0,1/2)^4+\vartheta_2(0,1/2)^4-1)/24$ and $2S_2 = S_1-1+\psi_{1/2}(i\pi/\log{2})/\log{2}$, in terms of the Jacobi theta and q-digamma functions, so I think there's a good chance there's not a simple form for $m>0$. $\endgroup$ – Zander Oct 21 '14 at 22:43

I think your observation helps if you iterate it more.

$$ \begin{align} S_m&=2-m\sum_{n=0}^\infty\frac{n}{2^n(2^n+m)}\\ &=2-m\sum_{n=0}^\infty \frac{n}{2^n}\left(\frac{1}{2^n}-\frac{m}{2^n(2^n+m)}\right)\\ &=2-m\sum_{n=0}^\infty \frac{n}{4^n}+m^2\sum_{n=0}^\infty\frac{n}{4^n(2^n+m)}\\ &=2-\frac49m+m^2\sum_{n=0}^\infty\frac{n}{4^n(2^n+m)}\\ \end{align}$$

Repeat this way, and brushing aside some convergence questions, we can write a power series in $m$:$$S_m=\sum_{n=0}^{\infty}(-1)^n\frac{2^{n+1}}{(2^{n+1}-1)^2}m^n$$ I'm not sure where you go from here, but you have an alternating power series now. Because it's alternating, it may be a quicker thing to use to get decimal approximations.


Nope! This power series doesn't converge for $m\geq2$. However, with $m=2$, it's partial sums oscillate between two values, the average of which appears to be $S_2$, so that's interesting.

As a power series, this converges for $m\in(-2,2)$. You could at least use this to study $S_m$ for (the mostly non-integer) $m$ in $(-2,2)$.

I examined a graph of the power series on $(-2,2)$, and it looked similar to functions of the form $\frac{2\cdot2^r}{(m+2)^r}$. Experimenting, using an $r$ in the neighborhood of $0.455$ gives $\frac{2\cdot2^r}{(m+2)^r}$ that matches the power series fairly well, and appears to give a decent approximation for $S_m$. So perhaps you can more rigorously find an $r\approx0.455$ such that $S_m$ is asymptotically proportional to $\frac{1}{(m+2)^r}$.

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    $\begingroup$ @Numberlover $m$ was in $(-2,2)$ while looking at a graph of the power series, in order to get a feel for what elementary function it resembled. But then I'm talking about general $m$. In other words, it looks like $S_m$ may be asymptotic to $\frac{c}{(m+2)^r}$ for some $r$. Consider $\{S_m\}\approx\{2, 1.67, 1.47, 1.32, 1.22, 1.13, 1.06,\ldots\}$ and $\left\{\frac{2\cdot2^{0.459}}{(2+m)^{0.459}}\right\}\approx\{2, 1.66, 1.45, 1.31, 1.21, 1.13, 1.06,\ldots\}$. $\endgroup$ – alex.jordan Oct 25 '14 at 18:26
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    $\begingroup$ Going back to the definition of $S_m$, it may be possible to prove such an asymptotic relationship. Or then again maybe not. Not much of what I wrote down is rigorous at all. $\endgroup$ – alex.jordan Oct 25 '14 at 18:30
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    $\begingroup$ @Numberlover Just to see what would happen, I did a regression on the $S_m$ versus $m+2$ using $m$ from $-1$ up to $11$. Specifically, I ran a linear regression on $\ln(S_m)$ versus $\ln(m+2)$, and the $r^2$-value is $0.9995\ldots$ with slope $-0.4647\ldots$ and intercept $1.0198\ldots$. What this means is that over $m$ in $\{-1,11\}$, the best power function fit to the $S_m$ data is $\frac{\exp(1.0198\ldots)}{(2+m)^{0.4647}}$. And the high $r^2$-value suggests the fit is very good. $\endgroup$ – alex.jordan Oct 26 '14 at 3:06
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    $\begingroup$ Again, just playing around, but I extended up to $m=20$, and $r^2$ reduced to $0.9987\ldots$ while the exponent increased to $0.4792\ldots$. It's at least a decent conjecture that you have something that is actually asymptotic to $c/\sqrt{m+2}$, since it would be "nice" if that exponent is approaching $1/2$. If you have some programming talent, you might explore going much higher than $20$ with this statistical investigation. $\endgroup$ – alex.jordan Oct 26 '14 at 3:14
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    $\begingroup$ I'm happy if I offered some insight. Your original observation mattered most though. On a whim I tried replacing the role that $2$ plays in the expression with other things like $3$, $4$, $5$, etc. With the conjecture that $2$ is leading to $c/(m+2)^{1/2}$, I was wondering if I'd see $c/(m+2)^{1/3}$, etc. But no such luck. At the small scale I am doing the statistics on, replacing the $2$ with $3,4,5,6$, and $7$ all lead to variations on what appears to be asymptotic to $c/(m+2)^{0.4\cdots}$. $\endgroup$ – alex.jordan Oct 26 '14 at 6:57

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