Switching the order of summations of a certain function

Question

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?

Answer 1

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.

Answer 2

$$\sum_{c=1}^{n-1}\,\sum_{k=c}^{n}=\sum_{k=1}^{n}\,\sum_{c=1}^{\min(k,n-1)}$$

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 $k$.

Answer 3

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 $$\lambda = -\sum_{k=1}^n \sum_{c=1}^{\min(k, n-1)} {k \choose c} \frac{(-1)^k}{k!} f^{k-c}U(-c,k-2c+1,-f)\phi(n,k)\\ = -\sum_{k=1}^n \frac{(-1)^k}{k!}\phi(n,k)\sum_{c=1}^{\min(k, n-1)} {k \choose c} f^{k-c}U(-c,k-2c+1,-f)\\ $$