Here is a prelim question that I have not been able to solve:
Show that if $a>1$, then $$ \sum_{n=1}^\infty n^{-1/2}a^{-n} = 1/\sqrt{\pi}\int_0^\infty \frac{y^{-1/2}}{ae^y-1}dy $$
Thanks so much, it has been driving me crazy!
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2There is something wrong with that. There is a "loose" $\sqrt n$ on the right hand side. – Pedro May 07 '13 at 00:38
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sorry, fixed it – Eager Student May 07 '13 at 00:41
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See here for the technique. – Mhenni Benghorbal May 07 '13 at 04:58
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We have $$\dfrac1{ae^y-1} = \dfrac1{ae^y} \dfrac1{1-\dfrac{e^{-y}}{a}} = \sum_{k=0}^{\infty} \dfrac{e^{-(k+1)y}}{a^{k+1}}$$ Hence, $$\int_0^{\infty} \dfrac{y^{-1/2}}{ae^y-1} dy = \sum_{k=0}^{\infty} \dfrac1{a^{k+1}} \int_0^{\infty} y^{-1/2} e^{-(k+1)y} dy$$Now use the $\Gamma$ function and finish it off. Equivalently, use the Gaussian integral, i.e., $$\int_0^{\infty} \exp(-x^2) dx = \dfrac{\sqrt{\pi}}2$$ Using above and some algebraic massaging gives us, $$\int_0^{\infty} y^{-1/2} e^{-(k+1)y} dy = \sqrt{\dfrac{\pi}{k+1}}$$ Hence, $$\dfrac1{\sqrt{\pi}}\int_0^{\infty} \dfrac{y^{-1/2}}{ae^y-1} dy = \sum_{k=1}^{\infty} \dfrac{a^{-k}}{\sqrt{k}} $$