n-1-1-2-n-2-

Question Number 98112 by M±th+et+s last updated on 11/Jun/20

\underset{n = 1}{\sum^{\infty}} \frac{1}{2^{n^{2}}}

Answered by maths mind last updated on 11/Jun/20

l e t s u s e

ϑ_{3} (z, τ) = \underset{n = - \infty}{\sum^{+ \infty}} q^{n^{2}} e^{2 i n z} T h e t a j a c o b i e F u n c t i o n

ϑ (0, \frac{1}{2}) = \underset{n = - \infty}{\sum^{+ \infty}} {(\frac{1}{2})}^{n^{2}} = 2 \sum_{n ⩾ 1} {(\frac{1}{2})}^{n^{2}} + 1

\sum_{n ⩾ 1} {(\frac{1}{2})}^{n^{2}} = \frac{1}{2} (ϑ (0, \frac{1}{2}) - 1)

Commented by M±th+et+s last updated on 11/Jun/20

t h a n k y o u s i r w e l l d o n e