Help finding the limit of a sum Hi I'm trying to find the following limit:
$$\lim_{n \rightarrow \infty} \frac{1}{n}  \sum_{j=1}^{ n } (1 - e^{\frac{-jt}{n}} )$$
expressed as a function of t. You may even be able to get it from Mathematica I don't have access to a copy at the moment.
Attempts made: tried to justify using L'hopital's rule, attempted to convert to integral. 

Look at $\frac{1}{n}  \sum_{j=1}^{ n } (1 - e^{\frac{-jt}{n}} )$.
$e^{\frac{-jt}{n}}
=\sum_{k=0}^{\infty} \frac{(-jt/n)^k}{k!}
$,
so
$1-e^{\frac{-jt}{n}}
=-\sum_{k=1}^{\infty} \frac{(-jt/n)^{k}}{k!}
$.
$\begin{align}
\sum_{j=1}^{ n } (1 - e^{\frac{-jt}{n}} )
&=\sum_{j=1}^{ n }\sum_{k=1}^{\infty} -\frac{(-jt/n)^{k}}{k!}\\
&=-\sum_{k=1}^{\infty}\sum_{j=1}^{ n } \frac{(-jt/n)^{k}}{k!}\\
&=-\sum_{k=1}^{\infty}\frac{(-t/n)^{k}}{k!}\sum_{j=1}^{ n } j^{k}\\
\end{align}
$
The innermost sum has been studied since Bernoulli,
and its value is known to be
$\sum_{j=1}^{ n } j^{k}
= \frac{n^{k+1}}{k+1}+O(n^k)
$.
The constant implied by the big-O is uniform, which justifies the liberties I am going to take.
What makes this work is that
when the term involving the big-O expression
is divided by $n$, it will go to $0$
as $n \to \infty$.
Using this,
the sum is
$\begin{align}
-\sum_{k=1}^{\infty}\frac{(-t/n)^{k}}{k!}
(\frac{n^{k+1}}{k+1}+O(n^k))
&=-n \sum_{k=1}^{\infty}\frac{(-t)^{k}}{(k+1)!}
+O(\sum_{k=1}^{\infty}\frac{(-t)^{k}}{k!})\\
&=\frac{-n}{-t} \sum_{k=1}^{\infty}\frac{(-t)^{k+1}}{(k+1)!}
+O(e^{-t}-1)\\
&=\frac{n}{t} (e^{-t}-1+t)
+O(1-e^{-t})\\
\end{align}
$
The final result is $1/n$ times this or
$\frac{1}{t} (e^{-t}-1+t)
+O((1-e^{-t})/n)
$.
As $n \to \infty$,
this becomes
$1-(1-e^{-t})/t
$.
Of course the Riemann sum to integral method is easier,
but I enjoyed doing this.
