Prove-or-disprove-the-foolowing-n-1-1-n-2-n-2-2-e-n-2-x-n-1-e-n-2-x-

Question Number 150539 by mathdanisur last updated on 13/Aug/21

Prove or disprove the foolowing :

\underset{\boldsymbol n = 1}{\sum^{\infty}} {(- 1)}^{\frac{\boldsymbol n^{2} + \boldsymbol n + 2}{2}} e^{- \boldsymbol {π n}^{2} \boldsymbol x} = \underset{\boldsymbol n = 1}{\sum^{\infty}} e^{- \boldsymbol {π n}^{2} \boldsymbol x}

Answered by Math_Freak last updated on 13/Aug/21

i just have to check if

{(- 1)}^{\frac{n^{2} + n + 2}{2}} will always be 1

\forall n : 1 ⩽ n ⩽ \infty, n \in Z

∴ \frac{n^{2} + n + 2}{2} should always be even

\frac{n^{2} + n}{2} + 1 should be even

\frac{n^{2} + n}{2} should be odd

\frac{n (n + 1)}{2} should be odd

n and n + 1 are conseutive numbers

∴ one of them must be even

if n is even and n + 1 is odd

then \frac{n}{2} (n + 1) can be even or odd

but if n is odd and n + 1 is even

then n (\frac{n + 1}{2}) can also be even or odd

∴ \frac{n (n + 1)}{2} is not always odd

and {(- 1)}^{\frac{n^{2} + n + 2}{2}} will not always be 1

\underset{\boldsymbol n = 1}{\sum^{\infty}} {(- 1)}^{\frac{\boldsymbol n^{2} + \boldsymbol n + 2}{2}} e^{- \boldsymbol {π n}^{2} \boldsymbol x} \neq \underset{\boldsymbol n = 1}{\sum^{\infty}} e^{- \boldsymbol {π n}^{2} \boldsymbol x}

Commented by mathdanisur last updated on 14/Aug/21

Thank you Ser

Thank you Ser