2
$\begingroup$

I'm almost certain that $$\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$$ converges. However, WebWork tells me that this is incorrect. I have been in that situation before but I obviously can't assume that I'm right and the computer is wrong based on that. I don't know how the system work and I'm not sure whether what I do is correct anymore.

By the $p$-test, I know $$3\sum_{n=1}^{\infty}\frac{1}{n^7}$$ converges. I also know that $\ln(n)$ grow very slowly so I use the comparison $\ln(n)\le Cn$ which I think will not change that fact that the sum converges such that $$\sum_{n=1}^\infty\frac{3\ln(n)}{n^7}\le \sum_{n=1}^\infty\frac{Cn}{n^7}= C\sum_{n=1}^\infty\frac 1{n^6}\lt\infty$$ which makes me quite certain that $\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$ is convergent.

I also checked that $$\begin{align} \lim_{\epsilon \to \infty}\int_1^\epsilon\frac{3\ln(x)}{x^7}\,dx & = 3\lim_{\epsilon \to \infty}\left(-\frac{\log (x)}{6 x^6}\bigg|_{1}^{\epsilon} +\frac{1}{6}\int_1^\epsilon \frac {dx}{x^7}\right)\\ & = 3\lim_{\epsilon \to \infty}\left( -\frac{\log (x)}{6 x^6}-\frac{1}{36 x^6}\right)\bigg|_{1}^{\epsilon}\\ & = \frac 1{12}\\ \end{align}$$

Can someone please shine some light on this for me?

$\endgroup$
3
  • $\begingroup$ Besides mixing $n$'s and $x$'s in your comparison, you are correct. The series converges. $\endgroup$ – carmichael561 Apr 4 '17 at 0:53
  • $\begingroup$ WolframAlpha agrees with you that this converges. $\endgroup$ – Mark Apr 4 '17 at 0:53
  • $\begingroup$ @carmichael561 Thank you I will edit that, I put that on the count of not being properly caffeinated... $\endgroup$ – user409521 Apr 4 '17 at 0:54
2
$\begingroup$

The series is convergent by p test and we may evaluate the “closed form”. The Riemann zeta function is defined as$$\zeta\left(s\right)=\sum_{n\geq1}\frac{1}{n^{s}},\,\textrm{Re}\left(s\right)>1.$$It is absolute convergent in the region $\textrm{Re}\left(s\right)>1$ so $$\zeta'\left(s\right)=\frac{d}{ds}\left(\sum_{n\geq1}\frac{1}{n^{s}}\right)=\sum_{n\geq1}\frac{d}{ds}\left(\frac{1}{n^{s}}\right)=-\sum_{n\geq1}\frac{\log\left(n\right)}{n^{s}} $$ hence $$S=3\sum_{n\geq1}\frac{\log\left(n\right)}{n^{7}}=\color{red}{-3\zeta'\left(7\right)}\approx0.0181.$$

$\endgroup$
2
  • 1
    $\begingroup$ Thanks for sparking my interest for Riemann's zeta function and giving me some insight. $\endgroup$ – user409521 Apr 4 '17 at 17:21
  • 1
    $\begingroup$ @InfiniteMonkey You're welcome. $\endgroup$ – Marco Cantarini Apr 4 '17 at 18:17
1
$\begingroup$

I would try the "Integral Test for Convergence" which says:

Consider an integer $N$ and a non-negative, continuous function $f$ defined on the unbounded interval $[N, ∞)$, on which it is monotone decreasing. Then the infinite series

$$\sum_N^{\infty} f(n)$$

converges to a real number if and only if the improper integral

$$\int_N^{\infty} f(x) dx$$

is finite.

Look at

$$\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$$

and the improper integral

$$\int _1^{\infty }\frac{3 \ln (n)}{n^7} = \frac{1}{12},$$ which is finite.

Thus, by the Integral Test for Convergence, you can say that the infinite series $\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$ converges.

$\endgroup$
1
  • $\begingroup$ The p test is more than enough. You don't need to solve that integral to know that the series converges, it is as immediate as $7 > 1$ $\endgroup$ – Ant Apr 4 '17 at 8:49
1
$\begingroup$

It really, really converges. The only common convergence tests I can think of that don't help are the root test and the alternating series test (the latter for the trivial reason that the summand is always positive).

Ratio test: $$\lim_{n \to \infty} \frac{\log(n+1)}{n \log(n)} = 0$$ which you can do by just "limit of a product is the product of the limits" on $\frac{\log(n+1)}{\log n} \times \frac{1}{n}$.

Comparison with $\frac{1}{n^2}$: $$\frac{\log n}{n^7} \leq \frac{1}{n^2}$$ if and only if $n^5 \geq \log n$, which is true for all positive $n$.

Integral test someone else has covered.

Cauchy condensation: it converges if and only if $$\sum_{n=0}^{\infty} \frac{n \log(2)}{2^{7n-n}}$$ does, but that is absurdly quickly convergent.

$\endgroup$
1
  • $\begingroup$ I was not aware of Cauchy condensation I'm reading about it now thanks! $\endgroup$ – user409521 Apr 4 '17 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy