Which of the following series is/are convergent? 
Which of the following series is/are convergent?


(a)$\sum_{n\ge2}\frac{1}{n\log n}$


(b)$\sum_{n\ge2}\frac{\log^2 n}{n^2}$


(c)$\sum_{n\ge2}\frac{1}{n\log^2 n}$


(d)$\sum_{n\ge2}\frac{\sqrt{n+1}-\sqrt{n-1}}{n}$

Solution:
(a) $\sum_{n\ge2}\frac{1}{n\log n}$ ~ $\int_{n=2}^{\infty}=+\infty$ and hence this series diverges.
(b)Using cauchy-condensation test we have: $\sum_{n\ge2}\frac{\log^2 n}{n^2}$ converges iff $\sum_{k\ge1}2^k\frac{\log^2 2^k}{2^{2k}}=\sum_{k\ge1}\frac {k^2\log^2 2}{2^{k}}$ converges. Since the series $\sum_{k\ge1}\frac {k^2}{2^{k}}$ converges using ratio test, we have that $\sum_{n\ge2}\frac{\log^2 n}{n^2}$ will converge.
(c)$\sum_{n\ge2}\frac{1}{n\log^2 n}$~$\int_{n=2}^{\infty}\frac{1}{n\log^2 n}$. The integral is finite and hence the series converges.
(d)$\sum_{n\ge2}\frac{\sqrt{n+1}-\sqrt{n-1}}{n}=\sum_{n\ge2}\frac{1}{n(\sqrt{n+1}+\sqrt{n-1})}$. Now, using Limit Comparison Test using the series $\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ will work as we get the limit $=\frac{1}{2}$ and hence the series is convergent.
I wanted to make sure, each of the solution/proof above are correct or not?
I wanted to know, if there is any other method other than cauchy-condensation test to solve the convergence/divergence of the series $\sum_{n\ge2}\frac{\log^2 n}{n^2}$ (given in part (b)).
Thanks for the help. Any valuable suggestions are welcome!
 A: Actually,
for any $c > 0$,
$\dfrac{\ln^c(x)}{x}
\to 0$
as $x \to \infty$.
To show this,
start with
$\dfrac{\ln(x)}{x}
\to 0$
as $x \to \infty$.
Then
$\dfrac{\ln(x^{1/c})}{x^{1/c}}
\to 0$
as $x \to \infty$.
But
$\begin{array}\\
\dfrac{\ln^c(x)}{x}
&=\left(\dfrac{\ln(x)}{x^{1/c}}\right)^c\\
&=\left(\dfrac{c\ln(x^{1/c})}{x^{1/c}}\right)^c\\
&=c^{1/c}\left(\dfrac{\ln(x^{1/c})}{x^{1/c}}\right)^c\\
&\to 0\\
\end{array}
$
A: 
"I wanted to know, if there is any other method other than Cauchy condensation test to solve the convergence/divergence of the series $\sum_{n=2}^\infty \frac{\log^2(n)}{n^2}$."


Using integration by parts with $\displaystyle u=\log^2(x)$ and $\displaystyle v-\frac1x$ followed by a second integration by parts with $\displaystyle u=2\log(x)$ and $\displaystyle v=-\frac1x$, it is easy to see that
$$ \int \frac{\log^2(x)}{x^2}\,dx=-\frac{\log^2(x)+2\log(x)+2}{x}+C\tag1$$
From $(1)$ we have
$$\begin{align}\lim_{L\to \infty}\int_2^L \frac{\log^2(x)}{x^2}\,dx&=\lim_{L\to\infty}\left(\frac{\log^2(2)+2\log(2)+2}{2}-\frac{\log^2(L)+2\log(L)+2}{L}\right)\\\\
&=\frac{\log^2(2)+2\log(2)+2}{2}
\end{align}$$
whence the integral test guarantees that the series $\displaystyle \sum_{n=2}^\infty \frac{\log^2(n)}{n^2}$ converges.


A: Your solutions are all correct.
For (b), to use Cauchy-condensation test I suppose you use the following fact :
$$\lim_{k\to\infty}\frac{\log^2 (2^k)}{(2^k)^2}/\frac{\log^2 (2^{k+1})}{(2^{k+1})^2}=4$$
This is a nice idea but maybe will take more time to write all down properly.
I suggest us to use this bound : $\log(n)=O(n^\epsilon)$ for any $\epsilon>0$ (this is called Big O notation : give two functions $f,g:\mathbb{R}^+\rightarrow\mathbb{R}^+$, we say that $f=O_{x\to\infty}(g)$ if there exist $C>0$ such that $f(x)<Cg(x)$ for $x$ big enough). Take say $\epsilon=\frac{1}{4}$ then the desired convergence follows.
