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

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

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

$$letsuse$$$${\vartheta }_3(z,\tau )=\underset{n=-\mathrm{\infty }}{\sum ^{+\mathrm{\infty }}}{q}^{n^2}{e}^{2inz}ThetajacobieFunction$$$$\vartheta (0,\frac{1}{2})=\underset{n=-\mathrm{\infty }}{\sum ^{+\mathrm{\infty }}}{\left(\frac{1}{2}\right)}^{n^2}=2\sum _{n⩾1}{\left(\frac{1}{2}\right)}^{n^2}+1$$$$\sum _{n⩾1}{\left(\frac{1}{2}\right)}^{n^2}=\frac{1}{2}(\vartheta (0,\frac{1}{2})-1)$$

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

$$thankyousirwelldone$$

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