Convergence of $\sum_{n=2}^\infty \frac{(\ln n)^3}{n^2}$

Question

I was trying to find if the series $\sum_{n=2}^\infty \frac{(\ln n)^3}{n^2}$ converges or diverges. But I couldn't solve the question and I looked at the solution in here. In that page, Limit Comparison Test is used with $\frac{(\ln n)^3}{n^2}$ and $\frac1{n^\frac32}$ and the limit found is 0. And it is conclued that the series converges. In my textbook, it says that this limit should be a finite number bigger than 0 in order to use Limit Comparison Test. So how does this work in this question, am I missing something?

Answer 1

In general if $a_n$ and $b_n$ are positve sequences such that $a_n/b_n\to 0$ and $$ \sum b_n<\infty $$ (i.e. converges to a finite value), then $\sum a_n<\infty$. To see this note that $$ a_n<b_n $$ for sufficiently large $n$ (say $n\geq l$ since $a_n/b_n\to 0$) whence $$ \sum_{n=l}^\infty a_n\leq \sum_{n=l}^\infty b_n<\infty $$

Answer 2

Since$$\lim_{n\to\infty}\frac{\frac{(\ln n)^3}{n^2}}{\frac{1}{n^{3/2}}}=0,$$you have that $\frac{(\ln n)^3}{n^2}\leqslant\frac{1}{n^{3/2}}$ if $n$ is large enough. So, since the series $\sum_{n=1}^\infty\frac1{n^{3/2}}$ converges, so does your series, by the direct comparison test.

Answer 3

It depends on how you state the limit comparison test.

Setup: Let $a_n,b_n>0$ and consider two series $\sum a_n,\sum b_n$.

One version: If $\lim a_n/b_n=c\in[0,\infty)$, then $\sum b_n$ converges implies $\sum a_n$ converges. Similarly if $\lim a_n/b_n=c\in(0,\infty]$, then $\sum a_n$ diverges implies $\sum b_n$ diverges.

Proof: If $\lim a_n/b_n=c<\infty$, then $\sup a_n/b_n=c'<\infty$ (because only finitely many terms are $>c+\varepsilon$), so comparison test yields the conclusion. The other statement is just the contrapositive form, swapping the roles of $a,b$.

The version in your textbook probably says: If $\lim a_n/b_n=c$ is finite and nonzero, then $\sum a_n,\sum b_n$ either converges together or diverges together. This is combining the two results above.