Finding convergence of a series using integral test The series:$$\sum_{n=1}^{\infty}\left(\frac{\ln(n)}{n}\right)^{2}$$
Question:
a) show that it converges
b) find the upper bound for the error in approximation $s\approx s_{n}$
Trial:
The section was about integral test, but the sequence$\left(\frac{\ln(n)}{n}\right)^{2}$ is not decreasing from [1,$\infty$]( it is increasing from [ 1,e ] ) so, I could not use the integral test. 
Other method:
I tried to find a sequence greater than$\left(\frac{\ln(n)}{n}\right)^{2}$so that it satisfies the condition for use of integral test( If I show using integral test that the new series is bounded then It could imply that our sequence is convergent since its bounded and decreasing).the problem was that I had trouble finding any function which could satisfy such conditions
b) this is understanding problem, is it asking me to find the exact sum or some upper bound , in any way how can I do this?
 A: Since $\log n\leq (n-1)^{\frac{2}{5}}$ for any $n\geq 1$,
$$0\leq \sum_{n\geq 1}\frac{\log^2 n}{n^2}\leq \sum_{n\geq 1}\frac{1}{n^{\frac{6}{5}}}$$
and the RHS is convergent by the $p$-test. Moreover,
$$ \sum_{n\geq 1}\frac{\log^2 n}{n^2}=\frac{d^2}{ds^2}\left.\sum_{n\geq 1}\frac{1}{n^s}\right|_{s=2} = \zeta''(2) = 1.9892802342989\ldots $$
A: For the convergence we can use for example, for $x
 $ sufficiently large (say $x\geq N
 $), $$\log\left(x\right)\leq x^{1/4}
 $$ hence $$\sum_{n\geq N}\frac{\log^{2}\left(n\right)}{n^{2}}\leq\sum_{n\geq N}\frac{1}{n^{3/2}}<\infty.
 $$ About the upper bound for the error, we can use the integral test $$\sum_{n\geq N}f\left(n\right)\leq f\left(N\right)+\int_{N}^{\infty}f\left(x\right)dx
 $$ and so in our case $$\sum_{n\geq1}\frac{\log^{2}\left(n\right)}{n^{2}}=\sum_{n=1}^{N}\frac{\log^{2}\left(n\right)}{n^{2}}+\sum_{n\geq N+1}\frac{\log^{2}\left(n\right)}{n^{2}}\leq
 $$ $$\leq\sum_{n=1}^{N}\frac{\log^{2}\left(n\right)}{n^{2}}+\frac{\log^{2}\left(N+1\right)}{\left(N+1\right)^{2}}+\int_{N+1}^{\infty}\frac{\log^{2}\left(x\right)}{x^{2}}dx
 $$ and the integral is, using the integration by parts, $$\int_{N+1}^{\infty}\frac{\log^{2}\left(x\right)}{x^{2}}dx=\frac{\log^{2}\left(N+1\right)}{N+1}+2\int_{N+1}^{\infty}\frac{\log\left(x\right)}{x^{2}}dx=
 $$ $$=\frac{\log^{2}\left(N+1\right)}{N+1}+\frac{2\log\left(N+1\right)}{N+1}+2\int_{N+1}^{\infty}\frac{1}{x^{2}}dx=
 $$ $$\frac{\log^{2}\left(N+1\right)+2\log\left(N+1\right)+2}{N+1}.
 $$
A: To show the series converges using the integral test we simply integrate by parts twice with successive substitutions $u_1=(\log x)^2 $ and $v_1=x^{-2}$, and  $u_2=\log x $ and $v_2=x^{-1}$, to reveal
$$\begin{align}
\int_3^{\infty} \left(\frac{\log x}{x}\right)^2\,dx&=-\left.\left(\frac{(\log x)^2}{x}\right)\right|_{3}^{\infty}+2\int_3^{\infty} \frac{\log x}{x^2}\,dx\\\\
&=\frac13 (\log(3))^2+2\int_3^{\infty} \frac{\log x}{x^2}\,dx\\\\
&=\frac13 (\log(3))^2-2\left.\left(\frac{\log x}{x}\right)\right|_{3}^{\infty}+2\int_3^{\infty} \frac{1}{x^2}\,dx\\\\
&=\frac13 (\log(3))^2+\frac23 \log (3)+\frac23
\end{align}$$
Thus, the series converges.

UPPER AND LOWER BOUNDS
To find an upper bound of the series using the integral test we use 
$$\begin{align}
\sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2&\le \left(\frac{\log 2}{2}\right)^2+\left(\frac{\log 3}{3}\right)^2+\int_3^{\infty}\left(\frac{\log x}{x}\right)^2\,dx\\\\
&=\left(\frac{\log 2}{2}\right)^2+\left(\frac{\log 3}{3}\right)^2+\frac13 (\log(3))^2+\frac23 \log (3)+\frac23\\\\
&\approx. 2.05560987295277
\end{align}$$
The lower bound is simply the upper bound less the third term $\left(\frac{\log 3}{3}\right)^2\approx. 0.134105440090287
$
Thus, we have
$$\bbox[5px,border:2px solid #C0A000]{
 \left(\frac{\log 2}{2}\right)^2+\frac13 (\log(3))^2+\frac23 \log (3)+\frac23 \le \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2}$$
$$\bbox[5px,border:2px solid #C0A000]{
\sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2\le \left(\frac{\log 2}{2}\right)^2+\left(\frac{\log 3}{3}\right)^2+\frac13 (\log(3))^2+\frac23 \log (3)+\frac23}$$
$$\bbox[5px,border:2px solid #C0A000]{
1.92150443286247
\le \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2\le 2.05560987295278}
$$
