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?

To estimate the sum , we first consider even $n$ and note the following: 


*

*$c = \frac{(1+e)}{2} \approx 1.85914\dots < e$, and therefore, the last term of the sum for $k=n$ can be ignored as $\binom{n}{n}\left(\frac{c}{e}\right)^n \rightarrow 0$ for $n \rightarrow \infty$. 

*$\binom{n}{k} = \binom{n}{n-k}$, and therefore, both terms $\binom{n}{k}\left(c \frac{k}{n}\right)^k$ and $\binom{n}{k}\left(c \frac{n-k}{n}\right)^{n-k}$ have the same binomial coefficient in the sum. 

*The idea is to estimate the sum of both terms by an exponential function of the form $c \mapsto e^{m k + b}$, where $m$ and $b$ depend only on $n$ and $c$.

*Indeed, the log of the sum, the function $k \mapsto \log\left( \left(c\frac{k}{n}\right)^k + \left(c\frac{n-k}{n}\right)^{n-k} \right)$, is strictly convex and on the interval $\left[1,\frac{n}{2}\right]$ it is upper-bounded by the linear function
$$ f(k) = \left(\frac{\log(4)}{n}-\log(2c)\right) k + n \log(c).$$

*Thus, $$\binom{n}{k}\left(c\frac{k}{n}\right)^k + \binom{n}{n-k}\left(c\frac{n-k}{n}\right)^{n-k} \leq \binom{n}{k} e^{f(k)}.$$


We can bound the sum by 
\begin{eqnarray}
\frac{1}{e^n} \sum_{k=1}^{ \frac{n}{2} } \binom{n}{k} e^{f(k)} & = & \frac{1}{e^n} \sum_{k=1}^{\frac{n}{2}} \binom{n}{k} 4^{\frac{k}{n}} c^n \left(\frac{1}{2c}\right)^k \\
& \leq &  4\frac{c^n \left(1+ \frac{1}{2c}\right)^n}{e^n} \\
& = & 4\left(\frac{\frac{1}{2} + c}{e}\right)^n  \approx 4\left(\frac{2.35914}{e}\right)^n.
\end{eqnarray}
Since $2.35914 < e$, we get exponential convergence to $0$. 
For odd $n$ the argument is essentially the same, except that we need to also consider the central term (for $k = \frac{n}{2}+1$) separately.
\begin{eqnarray}
\binom{n}{\frac{n}{2}+1} \left(c \frac{\frac{n}{2}+1}{2}\right)^{\frac{n}{2}+1} & \leq & 2^n  \left( \sqrt{c \left(\frac{1}{2} + \frac{1}{n}\right)}\right)^n \cdot c \left(\frac{1}{2} + \frac{1}{n}\right) \\
& = & \left( \sqrt{c \left(2 + \frac{4}{n}\right)}\right)^n \cdot c \left(\frac{1}{2} + \frac{1}{n}\right).
\end{eqnarray}
With $\sqrt{c \left(2 + \frac{4}{n}\right)} \approx 1.92828 < e$, the result follows for odd $n$ as well. 
Note on generality:
So basically, if you want to estimate the asymptotics of 
$$ \frac{1}{x^n} \sum_{k=1}^n \binom{n}{k} \left(c\frac{k}{n}\right)^k, $$
the limit is guaranteed to be $0$ as long as $c \in [0,x-0.5)$.
