Suppose $\sum_{n=1}^{\infty} x_n$ converges, and that $\vert x_n \vert$ is decreasing. Does $\sum_{n=1}^{\infty} \sin(x_n)$ converge?

Question

Suppose $x_n$ is a real sequence such that $\displaystyle\sum_{n=1}^{\infty} x_n$ converges. Suppose further that $\displaystyle \vert x_n \vert$ is decreasing. Then does $\displaystyle\sum_{n=1}^{\infty} \sin(x_n)$ converge also?

Without the condition that $\displaystyle \vert x_n \vert$ is decreasing, I can find a counter-example: Take a rapidly increasing sequence of positive integers $(k_n)_{n\in}\mathbb{N},$ and then let $x_{k_1}, x_{k_2},\ldots$ be an increasing sequence of negative numbers converging very slowly to $0,$ whilst for each $n\in\mathbb{N},\ $ for each $k_n<m<k_{n+1},\ x_{m}= \frac{-x_{k_n}}{k_{n+1}-k_n-1}.$ For then, for each $n\in\mathbb{N},\ \displaystyle\sum_{m=k_n}^{k_{n+1}-1} x_m = 0,\ $ however, for each $n\in\mathbb{N},\ \displaystyle\sum_{m=k_n}^{k_{n+1}-1} \sin(x_m)$ can be made to be $>\frac{1}{n+1},$ for example. The point of the condition, "$\displaystyle \vert x_n \vert$ is decreasing", is that is seems to disallow this type of counter-example.

Answer 1

We may assume (after removing finitely many terms) that $0 < |x_n| < 1$ for all $n\geq 1$. Let $\psi(x)=\frac{\sin{x}}{x}$ for $0<x<1$, it is a decreasing function.

Let $a_n=\psi(|x_n|)$: the sequence $(a_n)_{n \geq 1}$ is non-decreasing, goes to $1$, and one has $\sin{x_n}=x_na_n$.

Let $X_n=\sum_{p \geq n}{x_p}$ for any $p\geq 1$.

Then, for any $N \geq 1$, one has $$\sum_{n=1}^N{\sin{x_n}}=\sum_{n =1}^N{x_na_n}=\sum_{n=1}^N{(X_n-X_{n+1})a_n}=\sum_{n=1}^N{X_na_n}-\sum_{n=2}^{N+1}{X_na_{n-1}}=X_{N+1}a_N+X_1a_1+\sum_{n=2}^N{X_n(a_n-a_{n-1})}.$$

Because $(a_n)_n$ is nondecreasing, $\sum_{n \geq 2}{a_n-a_{n-1}}$ is absolutely convergent, while the sequence $(X_n)$ converges to zero hence is bounded.

Thus $\sum_{n\geq 2}{X_n(a_n-a_{n-1})}$ is absolutely convergent, therefore $\sum_{n\geq 1}{\sin{x_n}}$ is convergent.

Answer 2

Aphelli's proof is nice and simple. Here's a generalization to extend the result to other odd power series.

Notice that for any integer $k$, the series $\sum x_n^{2k+1}$ converges. Indeed, this is again the Abel test with $x_n^{2k+1} = x_n x_n^{2k}$, using the assumption on $(|x_n|)$ to deduce that $x_n^{2k}$ is decreasing. More explicitely, let $S_N = \sum_{n=0}^N x_n$ and $S_N^{(2k+1)} = \sum_{n=0}^N x_n^{2k+1}$, we find: $$S_N^{(2k+1)} = S_N x_N^{2k} + \sum_{n=0}^{N-1} S_n (x_n^{2k} - x_{n+1}^{2k})$$ and the sum on the right converges absolutely. If $M > 0$ is an upper bound for the sequence $(S_n)$, we find: $$|S_N^{(2k+1)}| \le M \left[x_N^{2k} + \sum_{n=0}^{N-1} (x_n^{2k} - x_{n+1}^{2k})\right] = M x_0^{2k}.$$

Let's come back to $\sin$: $$\sum_{n = 0}^N \sin(x_n) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} S_N^{(2k+1)}$$ Using the power series expansion of $\sin$ which always holds (we can swap summations because one of the two sums is finite). Now we want to show that this expression admits a limit for $N \rightarrow +\infty$. This is exactly the discrete dominated convergence theorem, using the estimate above.

You can easily see that the argument only uses that $\sin$ expands as an odd power series. So the argument should hold for any odd function admitting a power series expansion near the origin.