Suppose there is a positive sequence $\{a_{n}\}$ such that $$\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}\frac{\sqrt{n}}{\log a_{n}}=0$$ How can we prove that $$\lim_{k\rightarrow\infty}\sum^{\infty}_{n=k}e^{\sqrt{n}}a_{n}=0$$
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Use $\log a_n<0$ so that $-\varepsilon\log a_n> \sqrt n$ and $e^{\sqrt n}a_n =e^{\sqrt n\log a_n}$. – Lutz Lehmann Feb 22 '14 at 05:51
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Can give detailed answers process. Explanation and justification. Thank you very much. – user130405 Feb 22 '14 at 06:07
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$$e^{\sqrt{n}}a_{n}\neq e^{\sqrt{n}a_{n}}$$ – user130405 Feb 22 '14 at 13:40
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I swear this question has been asked yesterday but can't find it. – user88595 Feb 22 '14 at 14:12
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You get that for some $N$ and $ε>0$ that for all $n\ge N$
$a_n<1$, so that $\ln a_n<0$ and $-ε<\frac{\sqrt n}{\ln a_n}<ε$
where the second inequality is trivial because of the signs. So that (edit: after correction of a stupid error)
$$-\frac{\sqrt n}ε>\ln a_n\iff a_n<e^{-\frac{\sqrt n}ε}\iff e^{\sqrt n}a_n<e^{-\sqrt n(\tfrac1ε-1)}$$
which could just be enough to prove convergence of the series. For instance by using that there are 2m+1 numbers between $m^2$ and $(m+1)^2$, i.e., using that
$$\sum_{n=m^2}^{m^2+2m}e^{-\sqrt n(\tfrac1ε-1)}\le (2m+1)e^{-m(\tfrac1ε-1)}$$
resulting in a condensed series as upper bound that has the classical structure of $\sum_m mq^m$.
Lutz Lehmann
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$$-\frac nε>\sqrt n\ln a_n\iff e^{\sqrt n}a_n<e^{-\frac nε}$$ is not true,because $$e^{\sqrt{n}}a_{n}\neq e^{\sqrt{n}a_{n}}$$ – user130405 Feb 22 '14 at 13:32
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You are right, that was stupid. Now it should be correct. Still not very elegant. – Lutz Lehmann Feb 22 '14 at 14:08
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Let $\epsilon=\frac{1}{2}$, and $e^{-\sqrt{n}(\frac{1}{\epsilon}-1)}=e^{-\sqrt{n}}$,then $$\lim_{k=\infty}\sum_{n=k}^{\infty}e^{-\sqrt{n}}=0\Leftrightarrow \sum_{n=1}^{\infty}e^{-\sqrt{n}} \quad\text{converge}$$ using Integral criterion $$\int_{1}^{\infty}e^{-\sqrt{x}}dx=4e^{-1}$$ – user130405 Feb 23 '14 at 04:32