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What test should i apply for testing the convergence/divergence of $$\sum_{n=1}^{\infty} \sqrt{\ln{n}\cdot e^{-\sqrt{n}}}$$

Help with hints will be appreciated. Thanks

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up vote 4 down vote accepted

Hint: Lets use the comparison test. To understand what we should compare to, lets look at the logarithm of the terms in the series, as this will be easier to get a grasp on. Notice that $$\log\left(\sqrt{e^{-\sqrt{n}}\log n}\right)=-\frac{1}{2}\sqrt{n}+\frac{1}{2}\log\log n\leq -2\log n$$ for sufficiently large $n$. The last inequality follows since $2\log n+\frac{1}{2}\log \log n \leq \frac{1}{2}\sqrt{n}$ for large $n$, as the square root function grows much quicker than the logarithm. This tells us that $$\sqrt{e^{-\sqrt{n}}\log n}\leq\frac{1}{n^{2}}$$ for sufficiently large $n$.

Idea: The function $e^{n^\alpha}$ for $0<\alpha<1$ grows slower than any exponential function, $e^{Cn}$ but it grows faster than any power $n^\beta$.

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Simply superb. Thanks :) – user53627 Dec 18 '12 at 17:40
@DavidMitra: Thanks, fixed. – Eric Naslund Dec 18 '12 at 17:53

The $n$-th term is equal to $$\frac{\sqrt{\log n}}{e^{\sqrt{n}/2}}.$$ The intuition is that the bottom grows quite fast, while the top does not grow fast at all.

In particular, after a while the top is $\lt n$.

If we can show, for example, that after a while $e^{\sqrt{n}/2}\gt n^3$, then by comparison with $\sum \frac{1}{n^2}$ we will be finished.

So is it true that in the long run $e^{\sqrt{n}/2}\gt n^3$? Equivalently, is it true that in the long run $\sqrt{n}/2\gt 3\log n$? Sure, in fact $\lim_{n\to\infty}\dfrac{\log n}{\sqrt{n}}=0$, by L'Hospital's Rule, and in other ways.

Remark: A named test that works well here is the Cauchy Condensation Test. I believe that a more "hands on" confrontation with the decay rate of the $n$-th term is more informative.

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+1 for not actually using the Cauchy Condensation Test. (I never liked that test!) – Eric Naslund Dec 18 '12 at 18:32

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