Closed form solution for the series $\sum _{n=1} ^{\infty} \frac{1}{2^n+1}$

Asked Dec 10 '17 at 21:27

Active Dec 10 '17 at 22:10

Viewed 99 times

I'd like to start by thanking you for your time reading this question and your passion for mathematics.

It would be a pleasant surprise to me if there were a closed form expression for the sum:

$$\sum _{n=1} ^{\infty} \frac{1}{2^n+1}$$

Knowing well that it converges, as term by term, $\frac{1}{2^n+1}<\frac{1}{2^n}$, I searched for a taylor series expansion, I rewrote the expression as $\frac{1}{2^n}-\frac{1}{2^n \cdot (2^n+1)}=\frac {2^{-n}}{1+2^{-n}}=1-\frac {1}{1+2^{-n}}$ in hopes of something, telescoping, nth partial-sum-exposing, or otherwise, all to no avail.

So I ask more accomplished and brilliant minds their thoughts. Thank goodness for MSE.

edited Dec 10 '17 at 21:34

Dylan

16,575

asked Dec 10 '17 at 21:27

MaximusFastidiousIrreverence

1,735

Actually I don't have so much time to carefully analyse your series. My 1st idea is to look for the Riemann sum of some (i don't know yet which) integral. This would be a finite sum, say from 1 to n. Then pass with $n$ to infinity. – szw1710 Dec 10 '17 at 21:32
A closed form for this series involves the digamma function. – JJacquelin Dec 10 '17 at 21:38
So far I would need to familiarize myself with Fourier Expansions and Gamma Functions – MaximusFastidiousIrreverence Dec 10 '17 at 21:41
http://mathworld.wolfram.com/q-PolygammaFunction.html – Pierpaolo Vivo Dec 10 '17 at 21:49
There isn't going to be any simple solution. – Simply Beautiful Art Dec 10 '17 at 22:24
@SimplyBeautifulArt suspected so, I was just surprised that I couldn't find anything on this series anywhere when I searched for one – MaximusFastidiousIrreverence Dec 10 '17 at 22:32

Closed form solution for the series $\sum _{n=1} ^{\infty} \frac{1}{2^n+1}$

0 Answers0