n-0-e-n-2-n-0-e-2n-2-

Question Number 125994 by Dwaipayan Shikari last updated on 16/Dec/20

\frac{\underset{n = 0}{\sum^{\infty}} e^{- n^{2}}}{\underset{n = 0}{\sum^{\infty}} e^{- 2 n^{2}}}

Answered by Olaf last updated on 16/Dec/20

Jacobi Theta function :

ϑ (z, τ) = \underset{n = - \infty}{\sum^{n = + \infty}} \exp (π {i n}^{2} τ + 2 π i n z)

ϑ (0, \frac{i}{π}) = \underset{n = - \infty}{\sum^{n = + \infty}} e^{- n^{2}} = 2 \underset{n = 0}{\sum^{\infty}} e^{- n^{2}} - 1 (1)

ϑ (0, \frac{2 i}{π}) = \underset{n = - \infty}{\sum^{n = + \infty}} e^{- 2 n^{2}} = 2 \underset{n = 0}{\sum^{\infty}} e^{- 2 n^{2}} - 1 (2)

(1) and (2) : \frac{\underset{n = 0}{\sum^{\infty}} e^{- n^{2}}}{\underset{n = 0}{\sum^{\infty}} e^{- 2 n^{2}}} = \frac{ϑ (0, \frac{i}{π}) + 1}{ϑ (0, \frac{2 i}{π}) + 1}

Commented by Dwaipayan Shikari last updated on 16/Dec/20

T h a n k i n g y o u! B u t i s t h e r e a n y o t h e r w a y t o a p p r o x i m a t e ?

Commented by Olaf last updated on 16/Dec/20

There is no asymptotic formula

but a computation is possible .

See here :

https : / / hlabrande . fr / maths / pubs / theta . pdf

Commented by Dwaipayan Shikari last updated on 16/Dec/20

\underset{n ⩾ 0}{\sum^{\infty}} e^{- n^{2}} \sim \int_{0}^{\infty} e^{- x^{2}} d x + \lim_{x \to 0} \frac{e^{- x^{2}}}{2} + i \int_{0}^{\infty} \frac{e^{- {(i x)}^{2}} - e^{- {(- i x)}^{2}}}{e^{2 π x} - 1} d x

= \int_{0}^{\infty} e^{- x^{2}} d x + \frac{1}{2} + 0 \sim \frac{\sqrt{π} + 1}{2}

C a n w e u s e t h i s ?

A b e l p l a n a f o r m u l a

\underset{z ⩾ 0}{\sum^{\infty}} f (z) d z = \int_{0}^{\infty} f (z) d z + \frac{f (0)}{2} + i \int_{0}^{\infty} \frac{f (i t) - f (- i t)}{e^{2 π t} - 1} d t

Commented by Dwaipayan Shikari last updated on 16/Dec/20

T h a n k i n g y o u