Limit of series divided and outside factor

Question

What is the limit of $$\lim_{x \to \infty} \frac{\sum_{k=0}^{x} (k+1){e^k}}{x{e^x}}\quad \text{?}$$ I think that it may go to infinity but not sure and i don't know how to solve it.

Answer 1

The numerator is an arithmetico-geometric series.

\begin{align*} (1-e)\sum_{k=0}^{x} (k+1){e^k}&=\sum_{k=0}^{x} (k+1){e^k}-\sum_{k=0}^{x} (k+1){e^{k+1}}\\ &=\sum_{k=0}^{x} (k+1){e^k}-\sum_{k=1}^{x+1} k{e^{k}}\\ &=\sum_{k=0}^{x} (k+1){e^k}-\sum_{k=0}^{x} k{e^{k}}-(x+1)e^{x+1}\\ &=\sum_{k=0}^{x} {e^k}-(x+1)e^{x+1}\\ &=\frac{e^{x+1}-1} {e-1}-(x+1)e^{x+1}\\ \sum_{k=0}^{x} (k+1){e^k}&=\frac{(x+1)e^{x+1}}{e-1}-\frac{e^{x+1}-1} {(e-1)^2}\\ \frac{1}{xe^x}\sum_{k=0}^{x} (k+1){e^k}&=\frac{(x+1)e}{x(e-1)}-\frac{e-e^{-x}} {x(e-1)^2}\\ \lim_{x\to\infty}\frac{1}{xe^x}\sum_{k=0}^{x} (k+1){e^k}&=\frac{e}{e-1} \end{align*}

Answer 2

For $x$ a natural.

$$ \sum_{k=0}^{x}\left(k+1\right)e^{k}=\frac{1}{\left(e-1\right)^2}\left(\left(x+1\right)e^{x+2}-\left(x+2\right)e^{x+1}+1\right) $$ So $$\frac{\displaystyle \sum_{k=0}^{x}\left(k+1\right)e^{k}}{xe^x}=\frac{1}{\left(e-1\right)^2}\left(\frac{x+1}{x}e^{2}-\left(x+2\right)\frac{e}{x}+\frac{1}{xe^{x}}\right) $$ Hence, the limit exists and

$$\frac{\displaystyle \sum_{k=0}^{x}\left(k+1\right)e^{k}}{xe^x} \underset{x \rightarrow +\infty}{\rightarrow}\frac{e^2-e}{\left(e-1\right)^2}=\frac{e}{e-1} \approx 1.58197$$