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. 
$$\sum_{k=2}^\infty \frac{1}{\sqrt k (\ln k)^2} $$
 A: 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,
$$\ln k=\lim_{\alpha\to0}\frac{k^\alpha-1}\alpha.$$
A: 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.
A: You can prove the following: $\left(\ln k\right)^2 < \sqrt[3]{k}$ 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, $k^{1/6} > \ln k \implies \ln k - k^{1/6} < 0, k \ge k_0$. 
A: 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.
A: 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}
$
A: At the original time that I posted this, many of these answers did not make complete sense to me, so I figured that I would post my own answer to this question that would have made sense to me back then.
We will solve the general case, when $q \in (0,1)$ and $p>1$. To solve this problem, we will have to use both the direct comparison test and the integral test.
Let $\displaystyle a_k = \frac{1}{k^q (\ln k)^p}$ and $\displaystyle b_k = \frac{1}{k\ln k}$. Arguably, the hardest part of this problem is understanding that for sufficiently large $k$, $a_k > b_k$. To prove this, we take the limit of their quotient.
$$\lim_{k \to \infty} \frac{k \ln k}{k^q (\ln k)^p} = \lim_{k \to \infty} \frac{k^{1-q}}{(\ln k)^{p-1}}$$
Because $k^\alpha \gg (\ln k)^\beta$ for any $\alpha,\beta>0$, we know that the above limit is equal to $\infty$. Therefore, there must exist an $N$ such that for $k>N$, $k\ln k > k^q (\ln k)^p$. Thus, we conclude that for sufficiently large $k$,
$$k \ln k > k^q (\ln k)^p \iff \frac{1}{k\ln k} < \frac{1}{k^q (\ln k)^p} \iff b_k < a_k$$
Now that we have that out of the way, if we can prove that $\sum_{k=2}^\infty b_k$ diverges, then $\sum_{k=2}^\infty a_k$ must diverge as well by the direct comparison test.
To prove that $\sum_{k=2}^\infty b_k$ diverges, we use the integral test.
$$\begin{align*} 
\int_2^\infty \frac{1}{x \ln x} dx &= \lim_{b \to \infty} \int_2^b \frac{1}{x \ln x}dx, \qquad u = \ln x, \quad du = \frac{1}{x}dx \\
&=\lim_{b \to \infty} \int_{\ln 2}^{\ln b} \frac{1}{u}du \\
&= \lim_{b \to \infty} \Big[\ln|u|\Big]_{\ln2}^{\ln b}\\
&= \lim_{b \to \infty} \Big(\ln\big|\ln b\big| - \ln\big|\ln2\big|\Big)\\
&=\infty
\end{align*}$$
Since the improper integral diverges, the associated sum $\sum b_k$ must diverge as well from the integral test. Therefore, we may conclude that $\displaystyle \sum_{k=2}^\infty \frac{1}{k^q (\ln k)^p}$ diverges for $q\in (0,1)$ and $p >1$.
