Show the $\sum_{k=1}^{\infty}1/k(\log k)^{1+\epsilon}<\infty.$

Question

I am working on an exercise, at the end of which I need to show

$$\sum_{k=1}^{\infty}\dfrac{1}{k(\log k)^{1+\epsilon}}<\infty,\ \text{for}\ \epsilon>0\ \text{fixed}.$$

It is natural to try to get an upper bound of this series which converges, and as $k\geq 1$, we must have $$k(\log k)^{1+\epsilon}\geq (\log k)^{1+\epsilon},$$ and thus $$\sum_{k=1}^{\infty}\dfrac{1}{k(\log k)^{1+\epsilon}}\leq \sum_{k=1}^{\infty}\dfrac{1}{(\log k)^{1+\epsilon}},$$ but how could I show the latter series converges? does this series even make sense (since we have singularity at $k=1$)?

Thank you!

The latter series does not converge. In fact $\sum \frac{\mathrm{d}k}{\ln^u k}$ diverges for all $u$. For $u > 1$, by comparison with $\sum k^{-1}$. — Eric Towers, Dec 13 '19 at 23:04
@EricTowers oh.. okay.. then how should I show the original series converges? — JacobsonRadical, Dec 13 '19 at 23:05
The original series has the same problem with (not) being defined at $k = 1$. — Eric Towers, Dec 13 '19 at 23:06
@EricTowers yes I think the exercise itself has some flaw in it.... I would just argue for $k>2$. — JacobsonRadical, Dec 13 '19 at 23:20
@angryavian mostly yes. Another problem with this exercise is the $k=1$ term not defined, but perhaps it is the exercise itself's problem.... — JacobsonRadical, Dec 13 '19 at 23:21

score 2 · Accepted Answer · answered Dec 13 '19 at 23:12

For $L > 1$, $\frac{1}{k \ln^{1+\varepsilon}k}$ is a positive, monotonically decreasing function on $[L,\infty)$, so we may apply the integral test. Using the substitution $u = \ln k$,

$$ \int_L^\infty \frac{\mathrm{d}k}{k \ln^{1+\varepsilon}k} = \frac{1}{\varepsilon \ln^{\varepsilon} L} $$

is finite, so $$ \sum_2^\infty \frac{1}{k \ln^{1+\varepsilon}k} $$ converges.

Show the $\sum_{k=1}^{\infty}1/k(\log k)^{1+\epsilon}<\infty.$

1 Answers1