How to use the limit comparison test to prove series convergence

Expert intro: A clean way to handle positive-term series is to compare them with something you already know converges. Here the key move is to squeeze the smaller series underneath a convergent one.

Detailed walkthrough

(a) Why $\Sigma a_n$ converges

Assume $a_n>0$, $b_n>0$, and $\sum b_n$ converges. We are given

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

That limit means that for large $n$, the ratio $a_n/b_n$ is tiny. In particular, there exists some $N$ such that for every $n\ge N$,

$$\frac{a_n}{b_n}<1,$$

so

$$a_n<b_n \qquad (n\ge N).$$

Because the terms are positive, we can compare tails:

$$\sum_{n=N}^{\infty} a_n \le \sum_{n=N}^{\infty} b_n.$$

The right-hand side converges, so the left-hand side also converges by the comparison test. Adding the finite initial part $\sum_{n=1}^{N-1} a_n$ does not change convergence, hence $\sum a_n$ converges.

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

Take

$$b_n=\frac{1}{n^2}.$$

Since $\sum \frac{1}{n^2}$ converges, it is enough to show

$$\lim_{n\to\infty}\frac{\frac{\ln n}{n^3}}{1/n^2} =\lim_{n\to\infty}\frac{\ln n}{n}=0.$$

That limit is standard, so by part (a),

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

converges.

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

Use the fact that for small $x$,

$$1-\cos x \sim \frac{x^2}{2}.$$

A simple comparison is even enough: for all real $x$,

$$1-\cos x \le \frac{x^2}{2}.$$

With $x=\frac{1}{n^2}$,

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

Now compare with the convergent $p$-series $\sum \frac{1}{n^4}$. So

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

converges.

If you want to phrase part (b) strictly using part (a), you can also note that

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

so the series behaves like $\sum \frac{1}{n^4}$.

💡 Pitfall guide

A common slip is forgetting that part (a) needs positive terms. That matters because the comparison argument is based on inequalities between tails. Another easy mistake is choosing a comparison series that is not known to converge. For example, in (i) comparing with $\sum 1/n$ would be too weak; $\sum 1/n^2$ is the right size. In (ii), many students remember $1-\cos x\approx x$, but the correct small-angle behavior is quadratic, not linear.

🔄 Real-world variant

If the numerator in (i) were a slower-growing function, like $\ln\ln n$, the same idea would still work as long as you compare against a convergent $p$-series and the ratio goes to $0$. For (ii), if the angle were $1/n$, then $1-\cos(1/n)$ would behave like $1/(2n^2)$, so the series would still converge. If the angle were much larger, say $1/\sqrt n$, then the same expansion would give a term comparable to $1/n$, which would change the outcome.

🔍 Related terms

comparison test, limit comparison test, p-series