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So I want to show that $$\sum_{n=2}^{\infty}\displaystyle\frac{1}{n\ln(n)^{1.01}}$$ converges.

I suppose if $$\ln(n)>n^{0.01}$$then $$n\ln(n)>n^{1.01}$$ and so $$\displaystyle\frac{1}{n\ln(n)^{1.01}}<\displaystyle\frac{1}{n^{1.01}},$$ meaning that

$$\int\limits_{2}^{\infty}\displaystyle\frac{1}{n\ln(n)^{1.01}}\mathrm dn<\int\limits_{2}^{\infty}\displaystyle\frac{1}{n^{1.01}}\mathrm dn.$$

And since the RHS of the inequality converges by the $p$ test, the right hand term also converges, and so the series converges. Is this correct? If it is, is there a better (e.g. more straightforward/obvious/elegant...etc.) way of showing that the series converges?

Martin Sleziak
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tsp216
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  • no, should be $\int\limits_{2}^{\infty}\displaystyle\frac{1}{n\ln(n)^{1.01}}\mathrm dn=\int\limits_{2}^{\infty}\displaystyle\frac{1}{\ln(n)^{1.01}}\mathrm d\ln(n)$ – lion Dec 02 '17 at 20:01

1 Answers1

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Actually, for any given $\epsilon>0$, $\ln n < n^{\epsilon}$ for sufficiently large values of $n$.

However, you're on the right track that the integral test should be used here. If $p>1$, the series

$$\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$$

converges iff the integral

$$\int_{2}^{\infty} \frac{1}{x(\ln x)^p}\ dx$$

does. Can you take it from here, using the substitution $u=\ln x$?

Carl Schildkraut
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