Good morning! I'm having trouble with this problem...
how can we find the exact value of
$$\sum_{n=0}^\infty e^{-n^2}.$$
Thank you in advance to anyone who can help.
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7
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4$\begingroup$ Simply there is no closed form of this sum. $\endgroup$ – Crostul Mar 2 '18 at 18:07
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2$\begingroup$ look up jacobi theta functions $\endgroup$ – tired Mar 2 '18 at 18:09
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$\begingroup$ Wolfram Alpha says this here $$\frac{1}{2} \left(1+\vartheta _3\left(0,\frac{1}{e}\right)\right)$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 2 '18 at 18:14
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$\begingroup$ Are you sure the exponent is squared? If not, then the exact value is easy to find. $\endgroup$ – Clayton Mar 2 '18 at 18:17
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$\begingroup$ @Clayton, If the exponent non squared it's clear at all that is a progression geometric series such that convergent $\endgroup$ – zeraoulia rafik Mar 2 '18 at 18:28
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Hint: This may helping you and assured that no closed form exist only it is a special case for $\theta$ and $\psi$ functions
The sum :$\displaystyle\sum_{n=0}^\infty e^{-n^2}$ is a particular case of $\psi$ function at $\displaystyle x=\frac 1 \pi$ which is defined as :$\displaystyle\psi(x)=\sum_{n=0}^\infty e^{-n^2\pi x}.$ , For more information try to check (Edwards 2001, p. 15)and satisfies the functional equation :$\displaystyle\frac{1+2\psi(x)}{1+2\psi(x^{-1})}=\frac {1}{\sqrt{x}}$ , In your case you have :$\displaystyle \frac{1+2\psi(\pi)}{1+2\psi({\pi}^{-1})}=\sqrt{\pi}$ , and for more information about this you can see :(Jacobi 1828; Riemann 1859; Edwards 2001, p. 15) .
answered Mar 2 '18 at 18:47
