Convergence/Divergence of $\sum_{n=1}^{\infty} \sin(1/n)$

You need to determine convergence for $\sum_{n=1}^\infty\sin(1/n)$. The series diverges. The two hints below may guide you when trying to justify this.

Hint 1: $ \lim_{\theta\to0}\sin(\theta)/\theta=1$ and $1/n\to 0$ as $n\to\infty$.

Hint 2: $\sin(1/n)$ is positive. So you may attempt a limit comparison test.

Hint: $\sin(x) / x \to 1$ as $x \downarrow 0$.

We can prove this is divergent through direct comparison, though the way I know uses a Taylor Series expansion, which is not normally covered when students first see this problem. For others, this might shed some new light on the intuition that is lost from the limit comparison test. Sometimes it can feel like the limit comparison test just works "by magic," at least for me.

If you are unfamiliar with a Taylor Series expansion, I encourage you to research it. It becomes quite useful in most portions of applied math.

The first term of the Taylor Series can often hint at the convergence or divergence of a function, as you will see below. For this problem, expand $\sin{x}$ about $x=0$.

$$\sin{⁡x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$

If we truncated the Taylor series at the second term, we’d have the following positive Lagrange remainder: $$\sin{⁡x}=x-\frac{x^3}{3!}+f^{(5)}(c)\frac{x^5}{5!},$$ where $c$ is between $0$ and $x$. Note that $f^{(5)} (x)=\cos{⁡x}$, so we can replace $f^{(5)}(c)$ with $\cos⁡{(c)}$ from this point on.

Now, if we borrow this same Taylor expansion, but let $x=1/n$ for some positive integer n, we’d get $$\sin⁡{(\frac{1}{n})}=\frac{1}{n}-\frac{1}{6n^3}+\frac{\cos{⁡(c)}}{5!}\frac{1}{n^5}.$$

Like stated before, looking at the first term $1/n$ hints that this function might diverge. As a similar example, if we were instead observing the function $1-\cos{\frac{1}{n}}$, we'd find the first term to be $\frac{1}{n^2}$ which hints that it converges.

(This expansion is quite weird when you think about it. We built it centered around $x=0$ which would mean that with this substitution, we centered it around $\lim_{n→∞}⁡\frac{1}{n}=0$, an odd concept for Taylor expansions, but we’ll work with it.)

If you note that $1/n$ is always going to be positive and $0<\frac{1}{n}\leq1$, this means that $0<c<1$. As a result, $\cos{⁡(c)}>0$, regardless of $c$. The resulting remainder will be positive, so removing it will result in an under-approximation. To show this bring over the other two terms to the left, as seen below: $$\sin{⁡(\frac{1}{n})}-\frac{1}{n}+\frac{1}{6n^3}=\frac{\cos{⁡(c)}}{5!}\frac{1}{n^5}>0$$

$$\implies \sin{⁡(\frac{1}{n})}>\frac{1}{n}-\frac{1}{6n^3}$$ We can use this for our summation. Observing the corresponding series on the right hand, we have a divergent series: $$\sum_{n=1}^{\infty}{\frac{1}{n}-\frac{1}{6n^3}}=\sum_{n=1}^{\infty}{\frac{1}{n}}-\sum_{n=1}^{\infty}{\frac{1}{6n^3}}.$$ The second series above would be convergent, but the harmonic series $\sum_{n=1}^{\infty}{1/n}$ is famously divergent. By direct comparison test, because the above series is divergent and all terms of the above series are less than the corresponding terms of $$\sum_{n=1}^{\infty}{\sin⁡{(\frac{1}{n})}}$$ we have the following for our sine series $$\sum_{n=1}^{\infty}{\sin{⁡(\frac{1}{n})}}>\sum_{n=1}^{\infty}{\frac{1}{n}-\frac{1}{6n^3}}$$ meaning it too is divergent.