Convergence of the series $\sum_{n=1}^{\infty}\frac{3\ln(n)}{n^7}$; I say it converges, WebWork tells me this is incorrect. 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?
 A: 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 $$S=3\sum_{n\geq1}\frac{\log\left(n\right)}{n^{7}}=\color{red}{-3\zeta'\left(7\right)}\approx0.0181.$$
A: 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.
A: 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.
