How many non-negative integer solutions to $a_1+a_2+\cdots+a_k\leq n$ with given $n$ and $k$? 
I want to count the non-negative integer solutions to the following inequality:
  $$
a_1+\cdots+a_k\leq n
$$
  where $n$ and $k$ are given nonnegative integers.

I tried using stars and bars for each integer less than $n$ ($n-1,n-2,n-3\dots 3,2,1$), but that didn't help since I only got a sigma.
Btw the answer cannot involve a sigma or a ...
 A: You precise in a comment that the $a_i$ are non-negative integers.
In that case the number of solutions to the inequality corresponds to the number of  weak compositions of $m$
into exactly $k$ parts, with $0 \le m \le n$.
Now, the number of weak compositions of $m$ into $k$ parts is 
$$
\binom{m+k-1}{m}
$$
and the sum on $m$ gives
$$
\eqalign{
  & N(n) = \sum\limits_{m = 0}^n {\binom{m+k-1}{m}}  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,m\,\left( { \le \,n} \right)} {\binom{n-m}{n-m}\binom{m+k-1}{m}} 
 =  \binom{n+k}{n} \cr} 
$$
where:
 - the first step is to replace the summation bounds with the binomial $\binom{n-m}{n-m}$ that contains the upper bound implicitely,
while the lower bound is implicit in the original binomial;
 - the second step is to apply the "double convolution" , which in general reads
$$
\eqalign{
  & \sum\limits_{\left( { - m\, \le } \right)\,k\,\left( { \le \,n} \right)\,} 
 {\binom{r+k}{m+k} \binom{s-k}{n-k}}
  = \sum\limits_{\left( { - m\, \le } \right)\,k\,\left( { \le \,n} \right)\,} {\left( { - 1} \right)^{m + k} \binom{m-r-1}{m+k}
 \left( { - 1} \right)^{n - k} \binom{n-s-1}{n-k} }  =   \cr 
  &  = \left( { - 1} \right)^{m + n} \sum\limits_{\left( { - m\, \le } \right)\,k\,\left( { \le \,n} \right)\,} {\binom{m-r-1}{m+k} \binom{n-s-1}{n-k}}
  = \left( { - 1} \right)^{m + n} \binom{m+n-r-s-2}{m+n}  \cr 
  &  = \binom{r+s+1}{m+n}
\quad \left| \matrix{  \;r,s \in \mathbb C \hfill \cr   \,m,n \in \mathbb Z \hfill \cr}  \right. \cr} 
$$
applying the "upper negation" and the (standard) "convolution".
