Prove $\sum\limits_{n=1}^\infty ne^{-n^2} \leq \frac{3}{2e}$
Perhaps the mean value theorem for each term may work, but I haven't made any progress. I can't seem to reduce it to a geometric series either.
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$\begingroup$ Almost there: it is enough to invoke the derivative of a geometric series. $\endgroup$– Jack D'AurizioDec 18, 2017 at 21:07
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2 Answers
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Let $f(x) = x\exp(-x^2)$. Note that $f(x)$ is decreasing for $x \geq 1$. Therefore,
\begin{align*} \sum_{n=1}^{\infty}{f(n)} &= f(1) + \sum_{n=2}^{\infty}{f(n)} \\ &\leq \exp(-1) + \int_{1}^{\infty}f(x)dx \\ &= \exp(-1) + \int_{1}^{\infty}x\exp(-x^2)dx \\ &= \exp(-1) -2^{-1}\exp(-x^2)\bigg|_{1}^{\infty} \\ &= \frac{3}{2e} \end{align*}
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It is a pretty loose inequality:
$$ \sum_{n\geq 1}n e^{-n^2} = \frac{1}{e}+\sum_{n\geq 2} n e^{-n^2} \leq \frac{1}{e}+\sum_{n\geq 2} n e^{-2n}\leq \frac{6}{5e}\leq\frac{4}{9} $$ since $\sum_{n\geq 2}n x^n = \frac{x^2(2-x)}{(1-x)^2}$ for any $x\in[0,1)$.
answered Dec 18, 2017 at 21:06
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$\begingroup$ I wonder can we do it like that $\sum\limits_{n=1}^\infty ne^{-n^2}=e^{-1}+\sum\limits_{n=2}^\infty ne^{-n^2}\leq e^{-1}+\displaystyle\int_2^\infty xe^{-x^2}dx=e^{-1}+\frac{1}{2}e^{-4}<e^{-1}+\frac{1}{2}e^{-1}=\frac{3}{2e}.$ $\endgroup$– daulombDec 18, 2017 at 21:09
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$\begingroup$ I don't think it is true that $\sum_{n=2}^{\infty}{n\exp(-n^2)} \leq \int_{2}^{\infty}{x \exp(-x^2)dx}$. Check my solution for a similar idea. $\endgroup$– madprobDec 18, 2017 at 21:13
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$\begingroup$ @madroh: I didn't make sure that's why I asked:) $\endgroup$– daulombDec 18, 2017 at 21:15