Is $\sqrt{\pi}=2\sum_{k=0}^{\infty} \frac{e^{-k^2}-e^{-(k+1)^2}}{2k+1}$?

Asked Oct 27 '23 at 14:22

Active Oct 27 '23 at 14:31

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Is it true that $$ \sqrt{\frac{\pi}{4}}=\sum_{k=0}^{\infty} \frac{e^{-k^2}-e^{-(k+1)^2}}{2k+1} $$ Context: Attempting to find an easier proof for this estimate, when $x=1$.

My attempt: Leibniz's Formula gives $$ \frac{\pi}{4}=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1} $$ which is relatively closer to my expression. However, I am not sure how to rewrite this with the square root, if possible at all. Any ideas?

edited Oct 27 '23 at 14:31

asked Oct 27 '23 at 14:22

sam wolfe

3,465

Could this have something to do with the normal distribution? – Sai Mehta Oct 27 '23 at 14:29
Numerical evidence says "no". – emacs drives me nuts Oct 27 '23 at 14:39
@emacsdrivesmenuts I like that. How? – sam wolfe Oct 27 '23 at 14:44
Taylor expand $\sqrt(x)$ about a point of a=1 and then take $x=\frac{\pi}{4}$ and then equate that to the sum you are trying to show the equality to, and see that if their difference is 0 then it must be true. – Amy Skinner Oct 27 '23 at 14:47
WolframAlpha gives a value of $0.752298$ which isn't $\sqrt{\pi}/2$. – Luca Armstrong Oct 27 '23 at 14:48
https://www.wolframalpha.com/input?i=%5Csum_%7Bk%3D0%7D%5E%5Cinfty+%28%28e%5E%28-k%5E2%29-e%5E-%28%28k%2B1%29%5E2%29%29%2F%282k%2B1%29 – Sai Mehta Oct 27 '23 at 14:50
Just compute the first few terms in your preferred programming language.It turns out $\sum - \sqrt{\pi/4}\approx-0.134$. The series converges quickly ans is easy computationally, so no reason for a computation to be so far off of $0$. – emacs drives me nuts Oct 27 '23 at 14:50
I wonder if $\sum_{k=0}^{\infty} \frac{e^{-k^2}}{2k+1}$ is related to a theta function? – GEdgar Oct 27 '23 at 16:05
This is the same as $\sum _{k=-\infty }^{\infty } \frac{e^{-k^2}}{2 k+1}$ – Claude Leibovici Oct 28 '23 at 05:41

Is $\sqrt{\pi}=2\sum_{k=0}^{\infty} \frac{e^{-k^2}-e^{-(k+1)^2}}{2k+1}$?

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