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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?

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  • $\begingroup$ i don't understand your sum $\endgroup$ – Dr. Sonnhard Graubner Jul 24 '15 at 15:54
  • $\begingroup$ If the function is decreasing on $(e,\infty)$ then the sequence is decreasing on $[3,\infty)$. Removing a finite number of terms from the front end of a series won't change whether the series converges or not. $\endgroup$ – Alex Pavellas Jul 24 '15 at 15:56
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    $\begingroup$ btw, I assume you meant to use $n$ instead of $x$ inside the series. $\endgroup$ – Alex Pavellas Jul 24 '15 at 15:57
  • $\begingroup$ @Dr.SonnhardGraubner I made an edit $\endgroup$ – Socre Jul 24 '15 at 15:59
  • $\begingroup$ @user255545 yes, that infact is true but I have another problem, I had a difficult time integrating the function. If I knew how to integrate it your method is actually quite satisfying. $\endgroup$ – Socre Jul 24 '15 at 16:02
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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

$$\bbox[5px,border:2px solid #C0A000]{ \left(\frac{\log 2}{2}\right)^2+\frac13 (\log(3))^2+\frac23 \log (3)+\frac23 \le \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2}$$

$$\bbox[5px,border:2px solid #C0A000]{ \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2\le \left(\frac{\log 2}{2}\right)^2+\left(\frac{\log 3}{3}\right)^2+\frac13 (\log(3))^2+\frac23 \log (3)+\frac23}$$

$$\bbox[5px,border:2px solid #C0A000]{ 1.92150443286247 \le \sum_{n=1}^{\infty}\left(\frac{\log x}{x}\right)^2\le 2.05560987295278} $$

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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, $$ \sum_{n\geq 1}\frac{\log^2 n}{n^2}=\frac{d^2}{ds^2}\left.\sum_{n\geq 1}\frac{1}{n^s}\right|_{s=2} = \zeta''(2) = 1.9892802342989\ldots $$

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  • $\begingroup$ Why the exponent of $2/5$? $\endgroup$ – marty cohen Jul 24 '15 at 17:35
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    $\begingroup$ @martycohen: any constant slightly less than $\frac{1}{2}$ does the job, $\frac{2}{5}$ is just a nice number of that form. $\endgroup$ – Jack D'Aurizio Jul 24 '15 at 17:39
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    $\begingroup$ According to Wolfy, max{(log(log(x)))/(log(x-1))}~0.379831 at x~10.9352. So your choice of 2/5 is quite good. $\endgroup$ – marty cohen Jul 24 '15 at 17:47
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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, $$\int_{N+1}^{\infty}\frac{\log^{2}\left(x\right)}{x^{2}}dx=\frac{\log^{2}\left(N+1\right)}{N+1}+2\int_{N+1}^{\infty}\frac{\log\left(x\right)}{x^{2}}dx= $$ $$=\frac{\log^{2}\left(N+1\right)}{N+1}+\frac{2\log\left(N+1\right)}{N+1}+2\int_{N+1}^{\infty}\frac{1}{x^{2}}dx= $$ $$\frac{\log^{2}\left(N+1\right)+2\log\left(N+1\right)+2}{N+1}. $$

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