# Show the divergence of $\sum_{k=2}^\infty \frac{1}{\sqrt k (\ln k)^2}$

I'm attempting to understand how the following sum behaves for different values of $p$ and $q$: $$\sum_{k=2}^\infty \frac{1}{ k^q (\ln k)^p}$$I think I have figured out every case, except when $q \in (0,1)$ and $p>1.$ To determine what to do, I gave concrete values to $p$ and $q$. Namely, $q=\frac{1}{2}$, and $p=2$. But I still can't figure out how to prove the behavior of the series. Using Wolfram Alpha, I know that the series below diverges, but I don't know how to prove it. Does anyone have any insight to showing either the general case above or the more concrete case below? Thanks in advance.

$$\sum_{k=2}^\infty \frac{1}{\sqrt k (\ln k)^2}$$

• Can you apply the integral test? – Simply Beautiful Art Jul 20 '17 at 20:19
• I tried, but my if I set $u = \ln k$, and $du=1/k$, it doesn't quite work properly. Perhaps a different integration strategy would be better? – BSplitter Jul 20 '17 at 20:22
• That should be the correct approach. Could you show what you get after the substitution? – Simply Beautiful Art Jul 20 '17 at 20:25
• Are you referring to the general case or the specific one? – BSplitter Jul 20 '17 at 20:26
• Either is fine. – Simply Beautiful Art Jul 20 '17 at 20:27

## 5 Answers

Informally:

Actually, $\ln k$ grows slower than any positive power of $k$: for any $\alpha>0$ there is an $n$ such that

$$k>n\implies\ln k<k^\alpha$$

as, taking the derivative, $k^{-1}$ grows slower than $k^{\alpha-1}$.

Then you can replace $q$ by $q+\alpha<1$ and drop the $\ln$ factor, and the resulting series is bounded below by the harmonic one.

More about this:

In a way, $\ln k=k^0$. Indeed,

$$\ln k=\lim_{\alpha\to0}\frac{k^\alpha-1}\alpha.$$

For the general case, use the integral test for series. Observe that $$\sum \frac{1}{k^p\ln(k)^q} \approx \int_1^{\infty}\frac{dx}{x^p\ln(x)^q} = \int_0^{\infty}\frac{e^{-(p-1)u}}{u^q}du$$ It should be easier to deduce what values of $p$ and $q$ lead to convergence.

• I believe the exponents on this are flipped. Does that change the assessment at all? – BSplitter Jul 20 '17 at 20:35
• I switched the indices on accident but the analysis should be the same – Jonathan Davidson Jul 20 '17 at 20:38
• How would I consider integrating the last integral, with arbitrary powers? Integration by parts seems like the only option, but it looks awful. – BSplitter Jul 20 '17 at 20:46
• The integration by parts doesn't look to bad. If you need help with seemingly long integration by parts problems, look up the DI method on youtube, particularly on blackpenredpen's channel – Jonathan Davidson Jul 20 '17 at 20:50
• @BlakeSplitter ask yourself if the integrand approaches zero when $p <1$. – Simply Beautiful Art Jul 20 '17 at 21:48

You can prove the following: $\left(\ln k\right)^2 < \sqrt{k}$ for large enough $k \ge 2$. And this amounts to: $\ln k < k^{\frac{1}{6}}$. The ratio $\dfrac{k^{1/6}}{\ln k} \to +\infty$ as $k \to \infty$. Thus starting at $k_0$ on ward, $k^{1/6} > \ln k \implies \ln k - k^{1/6} < 0, k \ge k_0$.

From Cauchy's Condensation Test, the series $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}\log^2(n)}$ is convergent if anf only if the series $\sum_{n=1}^\infty \frac{2^{n/2}}{n^2\log^2(2)}$ converges. Since the latter series diverges (the general terms do not approach $0$), the series of interest diverges likewise.

The key fact is that $\dfrac{\ln x}{x^c} \to 0$ as $x \to \infty$ for any $c > 0$.

So show this, start with $\dfrac{\ln x}{x} \to 0$ as $x \to \infty$. Then $\dfrac{\ln x^c}{x^c} \to 0$ as $x \to \infty$. But $\dfrac{\ln x^c}{x^c} =\dfrac{c\ln x}{x^c} \to 0$.

Here is an easy way to show that $\dfrac{\ln x}{x} \to 0$.

$\ln(x) =\int_1^x \dfrac{dt}{t} \le\int_1^x \dfrac{dt}{t^{1/2}} =\dfrac{x^{1/2}-1}{1/2} \lt 2x^{1/2}$ so $\dfrac{\ln(x)}{x} \le \dfrac{2}{x^{1/2}} \to 0$.

You can also show it directly by a modification of this last proof.

For any $c > d > 0$ and $x > 1$,

$\begin{array}\\ \ln(x) &=\int_1^x \dfrac{dt}{t}\\ &\lt\int_1^x \dfrac{dt}{t^{1-d}} \qquad\text{since } t > t^{1-d}\\ &=\dfrac{x^{d}-1}{d}\\ &<\dfrac{x^{d}-1}{d}\\ \text{so}\\ \dfrac{\ln(x)}{x^c} &\le \dfrac{x^{d-c}}{d}\\ &= \dfrac{1}{dx^{c-d}}\\ &\to 0\\ \end{array}$