1

Approach 1

Evaluate $\sum_{\ell=1}^\infty \frac{1}{\ell(e^{2\pi \ell}-1)}$

The above link gave me the answer for a modified series, but if I were to adapt this to our case then it reduces to $\displaystyle\sum_{\ell\ge1}\frac1{\ell(e^{\ell\ln 2}-1)}$ and there's no imaginary number $i$ involved, so it becomes very a different case. So that didn't lead me anywhere.

Approach 2

I tried the geometric series approach by replacing $\frac1{2^\ell-1}$ with $\sum_{k\ge1}2^{-k\ell}$. But then it becomes the double summation:

$$\sum_{\ell\ge0}\sum_{k\ge1}\frac{2^{-k\ell}}\ell.$$

Upon swapping the summation we get the summation

$$\sum_{k\ge1}\ln(1-2^{-k}).$$

which is where I started to get to the summation in the question title above.

Show $\ln2 = \sum_\limits{n=1}^\infty\frac1{n2^n}$

The above link has the question closest to mine, for which I could find an existing answer underneath but I go around in circles or arrive at the above sum-of-logs.

Approach 3

Comparing to the appropriate integral leads to

$$\int_1^\infty\frac{dx}{x(2^x-1)}.$$

Substitution didn't do the trick either (on trying $u=2^x$ as well as $u=2^x-1$).

Approach 4

Is there function $f$ which has a Taylor expansion about $0$ with a radius of convergence move than 1, such that $\displaystyle f^{(\ell)}(0)=\frac{(\ell-1)!}{2^\ell-1}$?

If yes, that would do the trick, because then the summation is simply $f(1)$.

Approach 5

A good reference for an end-to-end computation would be nice. I couldn't find that either.

Karthik C
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  • Not quite. But I've added a line to the question (at the end). Answering that line will suffice. Equivalent to that last line, if the value of the summation in the link in your commenf above @Gary can be proven to be bigger than 1/4, that should do. The best available lower bound bigger than 1/4 would be appreciated! Thank you for your response! – Karthik C Mar 21 '22 at 05:34
  • What exactly would you like to bound from below by something larger than $1/4$? – Gary Mar 21 '22 at 05:58
  • ...the product in your first comment on this question. – Karthik C Mar 21 '22 at 06:37
  • I'm estimating the proportion of invertible matrices among all square matrices of size n, over the field with 2 elements. I wish to prove that over 25% of the matrices are invertible for large enough n. That's why the 1/4 (or 25%) lower bound. – Karthik C Mar 21 '22 at 06:41

1 Answers1

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I am answering the question in the comments. It has to be shown that $$ - \sum\limits_{k = 1}^\infty {\ln (1 - 2^{ - k} )} < \ln 4. $$ Note that for $0<x<\frac{1}{2}$, $$ - \ln (1 - x) = x + \frac{{x^2 }}{2} + x^3 \int_0^1 \!{\frac{{t^2 }}{{1 - xt}}dt} \le x + \frac{{x^2 }}{2} + x^3 \int_0^1\! {\frac{{t^2 }}{{1 - t/2}}dt} < x + \frac{{x^2 }}{2} + \frac{{2x^3 }}{3}. $$ Then $$ - \sum\limits_{k = 1}^\infty {\ln (1 - 2^{ - k} )} < \sum\limits_{k = 1}^\infty {2^{ - k} } + \frac{1}{2}\sum\limits_{k = 1}^\infty {4^{ - k} } + \frac{2}{3}\sum\limits_{k = 1}^\infty {8^{ - k} } = \frac{{53}}{{42}} < 1.3 < \ln 4. $$

Gary
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  • What is a good reference or a direction towards a proof of the first equality where the integral comes up? I want to back this up. I would certainly acknowledge you in my paper (in the pipeline) – Karthik C Mar 21 '22 at 08:10
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    Integrate both sides of $\frac{1}{{1 - s}} = 1 + s + \frac{s^2}{{1 - s}}$ from $0$ to $x$ and change the integration variable from $s$ to $t$ by $s=xt$. – Gary Mar 21 '22 at 08:54
  • Thanks Gary! That's comprehensive now. – Karthik C Mar 21 '22 at 08:58