Can we prove such an estimate $$\sum_{k=1}^\infty e^{-k^2t}\leq \frac{1}{2}\sqrt{\frac{\pi}{t}}?$$ I need it in my research....
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Actually the sum equals $\frac{1}{2}\sqrt{\frac{\pi}{t}}.$ – Mhenni Benghorbal Jun 20 '13 at 07:37
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1No, it does not. – XLDD Jun 22 '13 at 00:11
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HINT
Note that for a monotonically decreasing function, we have that $$\sum_{k=1}^{\infty} f(k) \leq \int_0^{\infty} f(x) dx$$ Move your mouse over the gray area for the solution.
In your case, $f(k) = e^{-k^2t}$. Hence, we have$$\sum_{k=1}^{\infty} e^{-k^2t} \leq \int_0^{\infty} e^{-x^2t} dx = \dfrac12 \sqrt{\dfrac{\pi}t}$$ from the following link: Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \dfrac{\sqrt \pi}{2}$
