Convergence of the Power Series of $\log(1+\sin x)$

Question

To find the first few terms in the power series of $\log(1+\sin x)$. The problem went onto expand as $\log(1+x)$ and then further expand by using the $\sin x$ series to get the answer.

My doubt is that the expansion of $\log(1+x)$ comes with a condition of $|x|<1$. But here $\sin x$ is not strictly < 1. So, is the problem wrong, or is it that I don't know something?

Answer 1

The power series for $\sin x$ converges for all $x$, so there is no concern there.

The power series of $\log(1 + x)$ converges only for $-1 < x \leq 1$ (notice the two different inequality signs). This implies that the power series of $\log(1 + \sin x)$ converges only for $-1 < \sin x \leq 1$. It is always true that $-1 \leq \sin x \leq 1$, so we should discard those $x$ for which $\sin x = -1$, namely $x = \frac{3\pi}{2} + 2k\pi$ for every $k \in \mathbb{Z}$.

If we let $B = \left\{\frac{3\pi}{2} + 2k\pi \mid k \in \mathbb{Z}\right\}$ ($B$ for "bad"), we can write that $\log(1 + \sin x)$ converges for $x \in \mathbb{R} \setminus B$.

Answer 2

The series for $\ln(1+x)$ converges for $x=1$. If you examine it, it passes the alternating series test. (The series does not of course converge for $x=-1$.)