Consider the series $\sum_{i\ge 1} (e^{(-1)^i\sin (\frac{1}{i})}-1)$.
For $x\to 0$, we have $e^x=1+x+O(x^2)$. Thus for $i\to \infty$ we have $$e^{(-1)^i\sin (\frac{1}{i})}-1=(-1)^i\sin(1/i)+ O(\sin^2 (1/i))$$
The two series $\sum_{i\ge 1}(-1)^i\sin(1/i)$ and $\sum_{i\ge 1} O(\sin^2 (1/i))$ converge (the former by the Leibniz test, the latter by the comparison test).
I was wondering would it be correct if I write $$\sum_{i\ge 1} (e^{(-1)^i\sin (\frac{1}{i})}-1)=\sum_{i\ge 1}(-1)^i\sin(1/i)+\sum_{i\ge 1} O(\sin^2 (1/i))$$ and conclude that $\sum_{i\ge 1} (e^{(-1)^i\sin (\frac{1}{i})}-1)$ converges? My concern is that the equality $$\sum_{i\ge 1} (e^{(-1)^i\sin (\frac{1}{i})}-1)=\sum_{i\ge 1} [(-1)^i\sin(1/i)+ O(\sin^2 (1/i))]$$ (and hence the previous equality) only holds as $i\to \infty$, but not in general. Do I have the right to write things like that and make conclusions about the convergence of the original series? If this use is incorrect, how do I write down my argument?
I would recommend writing $e^x=1+x+x^2H(x)$, where, by the uniqueness of Maclaurin polynomials, $H$ is a function bounded in some neighbourhood of $0$. This equality holds for any value of $x$.