Is $\displaystyle \sum_{t=1}^{\infty} e^{-\sqrt{t}}$ finite? This is not homework. Actually, a computer algebra package has told me it is finite, but I would like to see a proof. No approach worth trying has occurred to me - a hint would be appreciated...
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Hint: You can use the integral test. – Mhenni Benghorbal Aug 21 '13 at 02:42
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1This is a real answer not a trivial answer! Hints are regarded answers on this website! Based on what this answer was put as a comment? – Mhenni Benghorbal Aug 21 '13 at 03:16
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Note that $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots.$$
Put $x=\sqrt{t}$. We get that if $t\gt 0$ then $$e^{\sqrt{t}}\gt \frac{t^2}{4!}.$$
Thus the $t$-th term of our series is $\lt \dfrac{24}{t^2}$.
The series $\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent, and therefore so is $\displaystyle \sum_{n=1}^\infty \frac{24}{n^2}$. Thus by Comparison our series converges.
Remark: A very similar argument shows that if $\delta$ is any positive constant, then $\displaystyle\sum_{n=1}^\infty \frac{1}{e^{n^\delta}}$ converges.
André Nicolas
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Do you know that $$e^{-\sqrt{t}}=_\infty o\left(\frac{1}{t^2}\right)?$$