Estimating the Value of a Series — Question 8

*Question 8 Let $\displaystyle S=\sum_{n=1}^{\infty}ne^{-n}$ and $R_N=\sum_{n=N+1}^{\infty}ne^{-n}$. Use the Integral Test remainder theorem to give a two-sided bound for $R_N$.

*Question 8 – Solution

For $f(x)=xe^{-x}$, $$f'(x)=e^{-x}(1-x)\le0\qquad(x\ge1).$$ Thus $f$ is positive, continuous, and decreasing on the needed interval.

Integration by parts gives $$\int_A^{\infty}xe^{-x}\,dx=(A+1)e^{-A}.$$

$$\int_{N+1}^{\infty}xe^{-x}\,dx\le R_N\le\int_N^{\infty}xe^{-x}\,dx,$$ so $$\boxed{(N+2)e^{-(N+1)}\le R_N\le(N+1)e^{-N}}.$$ The upper bound is a decreasing integral majorant for the full omitted tail; no extra first-term adjustment is needed under this standard formulation.