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Question:

let $$\sum_{n=1}^{+\infty}\dfrac{1}{2^n-1}=E,\sum_{n=1}^{+\infty}\dfrac{1}{2^n+1}=F$$ where $E,F$ are constant,(in fact,$E$ is Erdős-Borwein Constant ),Find the sum $$f=\sum_{n=1}^{+\infty}(-1)^{n-1}\dfrac{2^{n+1}}{2^{2n}-1}$$

I tried this which I found in my texbook which seems relative, but Im not sure how to apply it to the problem: $$f=\sum_{n=1}^{+\infty}(-1)^{n-1}\left(\dfrac{1}{2^n+1}+\dfrac{1}{2^n-1}\right)=\sum_{n=1}^{+\infty}\dfrac{(-1)^{n-1}}{2^n-1}+\sum_{n=1}^{+\infty}\dfrac{(-1)^{n-1}}{2^n+1}$$ following can't try

math110
  • 93,304

2 Answers2

1

Let $d(m),d_1(m),d_0(m)$ be the number of divisors, odd divisors, even divisors of $m$. We have:

$$ F=\sum_{n\geq 1}\left(\frac{1}{2^n}-\frac{1}{2^{2n}}+\frac{1}{2^{3n}}-\ldots\right)=\sum_{m\geq 1}\frac{d_1(m)-d_0(m)}{2^m}=2\sum_{m\geq 1}\frac{d_1(m)}{2^m}-E $$ $$ E = \sum_{m\geq 1}\frac{d(m)}{2^m} $$

On the other hand:

$$\begin{eqnarray*}f=\sum_{n\geq 1}(-1)^{n+1}\frac{2^{n+1}}{(2^n-1)(2^n+1)}&=&\sum_{n\geq 1}(-1)^{n+1}\left(\frac{1}{2^n-1}+\frac{1}{2^n+1}\right)\\&=&2\sum_{n\geq 1}(-1)^{n+1}\left(\frac{1}{2^n}+\frac{1}{2^{3n}}+\frac{1}{2^{5n}}+\ldots\right)\\&=&2\sum_{m\geq 1}\frac{1}{2^m}\sum_{\substack{d\mid m \\ d\text{ odd}}}(-1)^{\frac{m}{d}+1}\\&=&2\sum_{\substack{m\geq 1\\m\text{ odd}}}\frac{d(m)}{2^m}-2\sum_{\substack{m\geq 1\\m\text{ even}}}\frac{d_1(m)}{2^m}\\&=&4\sum_{\substack{m\geq 1\\m\text{ odd}}}\frac{d(m)}{2^m}-2\sum_{m\geq 1}\frac{d_1(m)}{2^{m}}\end{eqnarray*} $$ and $$ E+F=2\sum_{m\geq 1}\frac{d_1(m)}{2^m}=2\sum_{\substack{m\geq 1\\m\text{ odd}}}d(m)\left(\frac{1}{2^m}+\frac{1}{2^{2m}}+\frac{1}{2^{4m}}+\frac{1}{2^{8m}}+\ldots\right) $$ so $$ f=\sum_{\substack{m\geq 1 \\ m\text{ odd}}}d(m)\left(\frac{2}{2^m}-\frac{2}{2^{2m}}-\frac{2}{2^{4m}}-\frac{2}{2^{8m}}-\frac{2}{2^{16m}}-\ldots\right) $$

$$ E=\sum_{\substack{m\geq 1 \\ m\text{ odd}}}d(m)\left(\frac{1}{2^m}+\frac{2}{2^{2m}}+\frac{3}{2^{4m}}+\frac{4}{2^{8m}}+\frac{5}{2^{16m}}+\ldots\right) $$

$$ F=\sum_{\substack{m\geq 1 \\ m\text{ odd}}}d(m)\left(\frac{2}{2^m}+\frac{0}{2^{2m}}-\frac{1}{2^{4m}}-\frac{2}{2^{8m}}-\frac{3}{2^{16m}}+\ldots\right) $$

and I do not see any way to write $f$ in terms of $E$ and $F$ only, without involving $\sum_{\substack{m\geq 1\\m\text{ odd}}}\frac{d(m)}{2^m}$ too.

In any case the numerical evaluation of such (Lambert) series is pretty simple: an acceleration technique is outlined here.

Jack D'Aurizio
  • 353,855
0

It's not possible to express $f$ only with $E$ and $F$ because of

$\displaystyle\sum_{n=1}^\infty (-1)^{n-1}\frac{2^{n+1}}{2^{2n}-1}=$

$\displaystyle -\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2^n-1} +\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2^n+1}+\frac{i}{2}\sum_{n=1}^\infty\frac{1}{2^n+i} -\frac{i}{2}\sum_{n=1}^\infty\frac{1}{2^n-i}$

$\displaystyle -\frac{1}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n+\sqrt{2}} +\frac{1}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n-\sqrt{2}}+\frac{i}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n+i\sqrt{2}} -\frac{i}{\sqrt{2}}\sum_{n=1}^\infty\frac{1}{2^n-i\sqrt{2}}$ .

You can define $\enspace\displaystyle g(x):=\prod_{n=1}^\infty (1+\frac{x}{2^n})\enspace$ (with the hope, that it exists a closed form for this product) and express $\enspace f\enspace$ by $\enspace\displaystyle \frac{d}{dx}\ln g(x)\enspace$ with $\enspace x\in\{-1;1;-i;i;-\sqrt{2};\sqrt{2};-i\sqrt{2};i\sqrt{2}\}$ .

user90369
  • 11,518