Convergence test using limit comparison with positive series

Key concept: This is a very standard convergence tool: compare a hard series with one you already know converges. The key is not the inequality alone, but the limit of the ratio.

Step by step

(a) Proof of the convergence claim

Assume $\sum a_n$ and $\sum b_n$ have positive terms, $\sum b_n$ converges, and

$\lim_{n\to\infty}\frac{a_n}{b_n}=0.$

Because the limit is $0$, for some $N$ we have

$0\le \frac{a_n}{b_n}<1 \quad \text{for all } n\ge N.$

So for $n\ge N$,

$0\le a_n\le b_n.$

Since $\sum b_n$ converges and all terms are positive, the tail $\sum_{n=N}^\infty b_n$ converges. By the direct comparison test,

$\sum_{n=N}^\infty a_n$

also converges. Adding the first finitely many terms does not affect convergence, so $\sum a_n$ converges.

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

Choose

$b_n=\frac1{n^2}.$

Then

$\frac{a_n}{b_n}=\frac{\ln n}{n^3}\cdot n^2=\frac{\ln n}{n} \to 0,$

because $\ln n$ grows much more slowly than $n$.

Since

$$$$

converges, part (a) implies

$\sum_{n=1}^{\infty}\frac{\ln n}{n^3}$ converges.

(b)(ii) $\sum_{n=1}^{\infty}\left(1-\cos\frac1{n^2}\right)$

Use the small-angle estimate

$1-\cos x \sim \frac{x^2}{2}\quad (x\to 0).$

With $x=\frac1{n^2}$,

$1-\cos\frac1{n^2} \sim \frac{1}{2n^4}.$

So a natural comparison series is

$b_n=\frac1{n^4},$

which converges.

More formally,

$\lim_{n\to\infty}\frac{1-\cos(1/n^2)}{1/n^4}=\frac12,$

so the given series behaves like a constant multiple of $\sum 1/n^4$, hence it converges.

Pitfall alert

A common mistake is to try to compare $\frac{\ln n}{n^3}$ with $\frac1{n^3}$ directly and then get stuck because the ratio is $\ln n$, not a finite limit. The trick is to compare with a slightly larger convergent series, such as $\frac1{n^2}$.

For the cosine series, do not split it as $\sum 1-\sum \cos(1/n^2)$; that is not a valid way to handle convergence here.

Try different conditions

If the cosine term were $1-\cos(1/n)$ instead, the same idea would still work, but the comparison would change to $\frac{1}{n^2}$ because $1-\cos x\sim x^2/2$. If the logarithm term were $\frac{\ln n}{n^p}$ with $p>2$, you could often compare it with $1/n^{p-1}$ or even $1/n^{p-\varepsilon}$ depending on the exponent.

Further reading

limit comparison test, direct comparison test, convergent positive series