How do you test the following series for convergence?: $$\sum_{n=2}^{\infty}\frac{1}{n^2\ln(n)}$$
What I've tried:
If we take the second sequence to be $1/n$ i.e $yⁿ=1/n$ and try to prove it by limit convergence test; but I dont know how can we prove that $1/n\geq 1/(n\ln(n))$
Justification for asking on a Q&A forum
To get an answer to my question, because I want to learn.
On
Well, the method you're trying isnt going to work. The limit comparison requires that you have a completely different series that you know converges. What youre doing is breaking your one and only series into term factors. I dont know where youre going with that. For example, if you want to use the limit comparison test then choose the series $\sum 1/n^2$ to compare it to:
Limit comparison:
If $\sum b_n = \sum \frac{1}{n^2}$ and $\sum a_n=\sum \frac{1}{n^2 \ln(n)}$ then evaluate the limit $\lim\frac{a_n}{b_n} = \lim \frac{n^2}{n^2 \ln(n)} = \lim \frac1{\ln(n)}=0$, which is finite.
Normally the limit must be positive but there is a caveat for the case of $0$. Since $\lim b_n\ne\infty$ we are still assured convergence.
But there are a multitude of other ways to prove this. Here are some other tests, below. Some you will learn in your calculus class and some you wont ever hear about unless you research them.
Direct comparison:
Observe that for $n\ge 3$, we have $\frac{1}{n^2\ln(n)}\le \frac{1}{n^2}$. So by direct comparison to the convergent power series, we have convergence.
Cauchy Condensation:
$\sum f(n) = \sum\frac{1}{n^2\ln(n)}$ converges if and only if $\sum 2^n f(2^n)=\sum \frac{2^n}{(2^n)^2\ln(2^n)} = \frac{1}{\ln(2)}\cdot\sum \frac{1}{n2^n}$ converges. Im ignoring indices since I dont care to evaluate the limits, but just to verify the convergence.
In any case, you can verify convergence of this new, modified series by the ratio test. If you have a keen eye and know your series', and pick the right starting index, you find that $\sum_{n=1}^\infty\frac1{n 2^n}=\ln(2)$
Cauchy Condensation is a good test for bounding a finite sum.
Ermakoff's Test
$\lim\frac{e^n f(e^n)}{f(n)}=\lim \frac{e^n/((e^n)^2\ln e^n)}{1/(n^2\ln n)}=\lim\frac{n\ln(n)}{e^n}=\lim\frac{1+\ln(n)}{e^n}=\lim\frac{1}{ne^n}=0<1$
Using two iterations of l'Hopitals rule. The limit superior is less than 1, therefore convergence. If limit inferior greater than 1, then divergence. Otherwise, indeterminate results.
Integral test
$\int_2^\infty f(x)\;dx = \int_2^\infty \frac{1}{x\ln(x)}\;dx\approx 0.3787$
Computer aided evaluation. It does not have a nice anti-derivative. But the integral is finite, so we have convergence.
Abel's Test
Make note that $\frac{1}{n^2\ln(n)}=\frac{1}{n^2}\cdot\frac{1}{\ln(n)}$ for two new sequences $\{1/n^2\}$ and $\{1/\ln(n)\}$.
Note that $\sum 1/n^2$ is convergent as per the p-series test and note that $\{1/\ln(n)\}$ is a bounded, monotone sequence for sufficiently large $n$.
Therefore $\sum\frac{1}{n^2\ln(n)}$ converges.
Rao's Logarithm test
Evaluate the limit: $$L=\lim_{n\to\infty} \frac{\ln(a_n)}{\ln(n)}=\lim\frac{\ln(\frac{1}{n^2\ln(n)})}{\ln(n)}=-2-\lim\frac{\ln(\ln(n))}{\ln(n)}$$
I dont know what that last funky limit evaluates to but it doesnt matter. Overall we are getting $L<-1$, thus convergence.
Yi-Fang Chang's Differential test
For $\sum a_n$ where $f(n)=a_n$ for $n\in\Bbb N$, we evaluate the limit of the continuous differential expression: $$L = \lim_{x\to\infty}x\frac{d}{dx}\ln(f(x))$$
In our case, when you evaluate the limit, we get $L=-2$ and since $L<-1$ convergence is implied.
I could go on to list a dozen or more other tests. Suffice it to say there are quite a few. You have options. Some are more useful or fruitful than others. It all depends on the series. Many fail entirely and some are just a lot of work for a series.
Hint. Note that for $n\geq 3$, $$\frac{1}{n^2\ln(n)}\leq \frac{1}{n^2}$$ and use the comparison test.