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?
 A: 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
$$ c < \frac{e^{2}(e^{2} - 1)}{e^{3} + 1}
\quad \Longrightarrow \quad S_{n} \leq \left( \frac{c}{e} + \frac{1}{e^{2}-e+1} \right)^{n} + \left( \frac{1}{e} + \frac{e^{2}-e+1}{e^{3}} c \right)^{n}
\xrightarrow[n\to\infty]{} 0. $$
A: To get an idea of the asymptotics, try using Stirling's approximation on the binomial coefficients:
$$\begin{align}\ln \left[{n \choose k} \left(\frac{c k}{n}\right)^k\right] &\sim \left[n \ln n - k \ln k - (n - k) \ln(n - k)\right] + \left[k \ln c + k \ln k - k \ln n\right] 
\\
&\sim (n-k) \ln\left(\frac{n}{n - k}\right) + k \ln c. \tag{1}\end{align}$$
Now, to see where the main "weight" of the summation is, i.e., for which values of $k$ the summands are the biggest, we just take the derivative with respect to $k$, set it equal to $0$, and solve for $k$:
$$1 + \ln \left(\frac{n-k}{n}\right) + \ln c = 0 \implies k = \left(1 - \frac{1}{ce}\right)n$$
Plugging this value of $k$ into $(1)$, we get that for this value of $k$ the summand becomes
$${n \choose k} \left(\frac{c k}{n}\right)^k \sim \exp\left(\left(\ln c + \frac{1}{e c}\right)n\right).$$
So for large $n$, the whole argument of the limit (including the term $e^n$) scales as
$$\exp\left(\left(\ln c + \frac{1}{e c} - 1\right) n\right).$$
Filling in $c = (e + 1)/2$ we get a negative exponent guaranteeing convergence of the order $e^{-0.182009 n} = 2^{-0.26258\dots n}$. We can verify these results numerically, e.g., for $n = 500$ the exact value is $3.35051\ldots \cdot 10^{-40}$ while the asymptotic approximation gives us $3.00019\ldots \cdot 10^{-40}$. The order of magnitute is exactly right, and only the constant is slightly off by about $10\%$.

If you would also like to know for which values of $c$ you get convergence, you just have to find out when the exponent is negative:
$$\ln c + \frac{1}{e c} - 1 < 0 \iff c(1 - \ln c) < \frac{1}{e}.$$
Substituting $c = e \cdot d$ will make the $1$ on the left hand side disappear:
$$d \ln d > \frac{-1}{e^2}.$$
This cannot be solved with elementary functions, but using the Lambert W function we get
$$d < e^{W(-1/e^2)} \implies c < e^{1 + W(-1/e^2)} = 2.31963\ldots.$$
A: We know that $\dbinom{n}{k} \le \left(\dfrac{e\,n}{k}\right)^k$. Therefore,
$$\sum_{k=1}^n \binom{n}{k} \left(c\frac{k}{n}\right)^k \le \sum_{k=1}^n \left(\dfrac{e\,n}{k}\right)^k \left(c\frac{k}{n}\right)^k = \sum_{k=1}^n (e\,c)^k = \frac{(e\,c)^{n+1}-1}{e\,c-1}$$
A: 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 $c \in [0,x-0.5)$.
