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: 
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\*}
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

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