Is this series G(1/n) convergent or divergent given G(x)?

Question

Suppose $G(x)=\int_0^x\sin{\left(e^s-1\right)}ds$

Does the series $\sum_{n=1}^{\infty}G(\frac{1}{n})$ converge or diverge?

I'm not sure how to go about solving this; however in our notes it says Taylor formula + $\int_{x_0}^x R_n(x-x_0)$ and I still can't make sense of it.

This is not homework, but a sample question for an upcoming midterm. Any ideas?

Answer 1

Hint

Suppose that you successively develop as a Taylor expansion $y=e^s-1$ and use the result for the development of $\sin(y)$, the integrand will be $$s+\frac{s^2}{2}-\frac{5 s^4}{24}-\frac{23 s^5}{120}+O\left(s^6\right)$$ Integrating between $0$ and $x$ will then give $$G(x) \simeq \frac{x^2}{2}+\frac{x^3}{6}-\frac{x^5}{24}-\frac{23 x^6}{720}+O\left(x^7\right)$$ Now replace $x$ by $\frac{1}{n}$ to obtain $$G(\frac {1}{n})=\frac{1}{2 n^2}+\frac{1}{6 n^3}-\frac{1}{24 n^5}-\frac{23}{720 n^6}$$

I am sure that you can take from here.