<s>[INST] Asymptotic behavior of $\sum_{k=1}^n \binom{n}{k} \left(\frac{ck}{n}\right)^k$ I am looking to show that 
$$\lim_{n \rightarrow \infty}\frac{1}{e^n}\sum_{k=1}^n \binom{n}{k} \left(\frac{ck}{n}\right)^k = 0. $$
In my application, $c = (e+1)/2 \approx 1.85914\ldots$. I have been looking all over the place, but I can't seem to find a closed form expression or good upper bound for the sum.
The obvious estimation
$$\sum_{k=1}^n \binom{n}{k} \left(c\frac{k}{n}\right)^k \leq \sum_{k=0}^n \binom{n}{k} \left(c\right)^k \leq \left(1+c\right)^n $$
won't do the trick, since $1+c=1+(1+e)/2 > e$. 
Any ideas?
 [/INST] I was also working with the inequality
$$ \left(1 - \frac{k}{n}\right)^{n-k} \leq \exp \left( -k + \tfrac{k^{2}}{n} \right) = \exp\left\{ -n \cdot \tfrac{k}{n} \left( 1 - \tfrac{k}{n} \right) \right\}.$$
Now note that $q(x) = x(1-x)$ satisfies $q(1-x) = q(x)$ and $q(x) \geq \frac{1}{2}x $ on $[0, \frac{1}{2}]$. Then we get
$$ 0 \leq k \leq \tfrac{1}{2}n \quad \Longrightarrow \quad \exp\left\{ -n \cdot \tfrac{k}{n} \left( 1 - \tfrac{k}{n} \right) \right\} = e^{-nq(k/n)} \leq e^{-k/2}. \tag{*} $$
To utilize this bound, we divide the sum into two part:
\begin{align*}
S_{n} := 
e^{-n} \sum_{k=0}^{n} \binom{n}{k} \left( \frac{k}{n} \right)^{k} c^{k}
&= e^{-n} \sum_{k=0}^{n} \binom{n}{k} \left( 1 - \frac{k}{n} \right)^{n-k} c^{n-k} \\
&\leq e^{-n} \sum_{k=0}^{n} \binom{n}{k} c^{n-k} e^{-nq(k/n)} \\
&\leq e^{-n} \sum_{k \leq \frac{n}{2}} \binom{n}{k} c^{n-k} e^{-nq(k/n)} + e^{-n} \sum_{\frac{n}{2} \leq k \leq n} \binom{n}{k} c^{n-k} e^{-nq(k/n)} \\
&= (c/e)^{n} \sum_{k \leq \frac{n}{2}} \binom{n}{k} c^{-k} e^{-nq(k/n)} + e^{-n} \sum_{k \leq \frac{n}{2}} \binom{n}{k} c^{k} e^{-nq(k/n)}.
\end{align*}
Here, in the last line, we applied the change of index $k \mapsto n-k$. Then we can use $\text{(*)}$ and we get the following crude bound:
\begin{align*}
S_{n}
&\leq (c/e)^{n} \sum_{k \leq \frac{n}{2}} \binom{n}{k} \left( \frac{1}{c\sqrt{e}} \right)^{k} + e^{-n} \sum_{k \leq \frac{n}{2}} \binom{n}{k} \left( \frac{c}{\sqrt{e}} \right)^{k} \\
&\leq \left\{ \frac{c}{e} \left( 1 + \frac{1}{c\sqrt{e}} \right) \right\}^{n} + \left\{ \frac{1}{e} \left( 1 + \frac{c}{\sqrt{e}} \right) \right\}^{n}.
\end{align*}
Now note that $c > 0$ satisfies the following condition
$$ \frac{c}{e} \left( 1 + \frac{1}{c\sqrt{e}} \right) < 1 \quad \text{and} \quad \frac{1}{e} \left( 1 + \frac{c}{\sqrt{e}} \right) < 1
\quad \Longleftrightarrow \quad c < e - \frac{1}{\sqrt{e}} \simeq 2.111751169. $$
Since your $c$ is less than $2$, the claim follows.

A slightly generally, you can introduce a parameter $\delta \in (0, 1)$. Then
\begin{align*}
\begin{array}{cl}
0 \leq k \leq \delta n
& \quad \Longrightarrow \quad \left( 1 - \tfrac{k}{n} \right)^{n-k} \leq e^{-(1-\delta)k}, \\
0 \leq k \leq (1-\delta) n
& \quad \Longrightarrow \quad \left( 1 - \tfrac{k}{n} \right)^{n-k} \leq e^{-\delta k}
\end{array}
\end{align*}
and our estimation is refined as
$$ S_{n} \leq \left( \frac{c}{e} + \frac{1}{e^{2-\delta}} \right)^{n} + \left( \frac{1}{e} + \frac{c}{e^{1+\delta}} \right)^{n}. $$
For both ratios to be less than 1, we must have $c < \min\{ e - e^{\delta-1}. e^{\delta}(e - 1) \}$. Maximizing this bound gives 
$$ e^{\delta} = \frac{1 + e^{-1}}{1 + e^{-3}}, $$
Therefore
$$ c < \frac{e^{2}(e^{2} - 1)}{e^{3} + 1}
\quad \Longrightarrow \quad S_{n} \leq \left( \frac{c}{e} + \frac{1}{e^{2}-e+1} \right)^{n} + \left( \frac{1}{e} + \frac{e^{2}-e+1}{e^{3}} c \right)^{n}
\xrightarrow[n\to\infty]{} 0. $$
</s>