Convergence and Divergence of Series — Question 8

*Question 8 Determine whether $\displaystyle\sum_{n=1}^{\infty}\left(1-\cos\frac1n\right)$ converges or diverges.

1. Verify that the terms approach zero.

2. Use $0\le1-\cos x\le x^2/2$ to obtain a direct comparison.

3. Confirm the quadratic scale by computing a limit-comparison ratio.

*Question 8 – Solution

Since $1/n\to0$ and cosine is continuous, $1-\cos(1/n)\to1-\cos0=0$. The necessary condition holds, so a convergence test is still needed.

The standard small-angle inequality $$0\le1-\cos x\le\frac{x^2}{2}$$ holds for every real $x$. Substituting $x=1/n$ yields $$0\le1-\cos(1/n)\le\frac1{2n^2}.$$ Both sides are nonnegative, and $\sum1/(2n^2)$ is a convergent $p$-series with $p=2>1$. The Direct Comparison Test therefore proves convergence.

The standard Taylor limit confirms more precisely that $$\lim_{n\to\infty}\frac{1-\cos(1/n)}{1/n^2} =\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12.$$ Thus the summand is asymptotic to $1/(2n^2)$. The finite positive ratio also satisfies the hypotheses of the Limit Comparison Test, independently confirming the conclusion.