$$S=\sum_{n=0}^{\infty}\frac1{e^n(n^2+1)}$$
This series converges because the general term goes to $0$ faster than $\dfrac1{n^2}$
I am asked to approximate the series with an error $R<10^{-3}$.
How can I do this? I know that
$$R=\left\lvert S-S_k\right\rvert=\left\lvert \sum_{n=0}^{\infty}\frac1{e^n(n^2+1)}-\sum_{n=0}^{k}\frac1{e^n(n^2+1)}\right\rvert=\sum_{n=k+1}^{\infty}\frac1{e^n(n^2+1)}$$ and that
$$\int_{k+1}^\infty \frac1{e^x(x^2+1)}\mathrm dx\le\sum_{n=k+1}^{\infty}\frac1{e^n(n^2+1)}\leq\int_{k}^\infty \frac1{e^x(x^2+1)}\mathrm dx$$
My attempt:
$$R=\sum_{n=k+1}^{\infty}\frac1{e^n(n^2+1)}\le\frac1{(k+1)^2+1}\sum_{n=k+1}^{\infty}\frac1{e^n}\underbrace\le_{(1)}\frac1{(k+1)^2+1}\int_k^\infty \frac1{e^x}\mathrm dx= \frac1{(k+1)^2+1}\frac1{e^k}<10^{-3}\quad (*)$$
$(1):$ as the series is decreasing, positive and continuous, I can switch to the integral.
$(*)$ is true for $k=4$.
Is this a right approach?
P.S. The exercise comes from Calculus Problems, 16.18, page 231.
An easily computable upper bound is
$$R =\sum_{n=k+1}^{\infty}\frac1{e^n(n^2+1)} \lt \frac1{(k+1)^2}\sum_{n=k+1}^{\infty}\frac1{e^n} =\frac1{(k+1)^2e^{k+1}(1-1/e)}$$