## Calculus: Early Transcendentals 8th Edition

We begin with the series: $$\sum_{k=1}^{\infty}ke^{-k}$$ Using the ratio test, we know that the series converges if the following limit converges: $$\lim_{k\to\infty}\frac{(k+1)e^{-(k+1)}}{ke^{-k}}$$ Algebraically manipulating the limit, we get: $$\lim_{k\to\infty}\frac{(k+1)e^{-(k+1)}}{ke^{-k}}\\ =\lim_{k\to\infty}\frac{(k+1)e^{-k-1}}{ke^{-k}}\\ =\lim_{k\to\infty}\frac{1}{e}\frac{(k+1)e^{-k}}{ke^{-k}}\\ =\lim_{k\to\infty}\frac{1}{e}\frac{k+1}{k}\\ =\frac{1}{e}$$ Since $\frac{1}{e}<1$, the series converges.