Switching the order of summations of a certain function I am looking to switch the order of the summations of the following function:
$$
\lambda = -\sum_{c=1}^{n-1} \sum_{k=c}^n {k \choose c} \frac{(-1)^k}{k!}  f^{k-c}U(-c,k-2c+1,-f)\phi(n,k)
$$
I don't know how to do this, can someone please teach me?
 A: This is what I do, and last did as recently as last week: draw a $ck$-coordinate system, and make a dot for each grid point that is in the sum. Then read it off.
A: $$\sum_{c=1}^{n-1}\,\sum_{k=c}^{n}=\sum_{k=1}^{n}\,\sum_{c=1}^{\min(k,n-1)}$$
The first sum is in a triangular grid, less the point $k=n$, $c=n$.  When summing first on $k$, $k$ starts at the variable index $c$ and ends at $n$. The outer sum in this case extends over the permissible values for $c$.
Now, when summing first on $c$, $c$ starts at $1$ and extends to the variable index $k$ (except when $k=n$ since $c\le n-1$) or $n-1$.  The outer sum then extends over the permissible values for $k$.
A: When I look at
$$\lambda = -\sum_{c=1}^{n-1} \sum_{k=c}^n {k \choose c} \frac{(-1)^k}{k!}  f^{k-c}U(-c,k-2c+1,-f)\phi(n,k)
$$
I see that
$1 \le c \le n-1$
and
$c \le k \le n$.
(Though I am suprised that
$c$ goes up to $n-1$, not $n$.)
Therefore
$1 \le k \le n$
and
$1 \le c \le k$
and $c \le n-1$.
The sums could then be rearranged as
$$\lambda = -\sum_{k=1}^n \sum_{c=1}^{\min(k, n-1)} 
{k \choose c} \frac{(-1)^k}{k!}  f^{k-c}U(-c,k-2c+1,-f)\phi(n,k)\\
= -\sum_{k=1}^n \frac{(-1)^k}{k!}\phi(n,k)\sum_{c=1}^{\min(k, n-1)} 
{k \choose c}   f^{k-c}U(-c,k-2c+1,-f)\\
$$
