# Sum dependant on another sum

I want to calculate the sum:

$$\sum_{k=1}^{1000} \frac{\lambda_1^k\exp(-\lambda_1)}{k!}(1-\sum_{n=0}^{k-1} \frac{1}{n!}exp(-\lambda_2 x)(\lambda_2 x)^n)$$

Wolfram Alpha seems to be having trouble with the second summation, I think the problem is that the stopping point of the second summation depends on the first summation. Is Wolfram Alpha able to solve this summation or is there different software I can use?

• In the second sum, factor out $\exp(-\lambda)$. The result doesn't have a particularly nice form and is known as the exponential sum function. – Simply Beautiful Art Dec 4 '17 at 15:19
• Ok, but that form still has summations dependant on a variable. Unless Wolfram Alpha has a function for the gamma function. – John Meighan Dec 4 '17 at 15:23
• My point was that I doubt there is a good closed form unless maybe rearranging the order of the sums does something. – Simply Beautiful Art Dec 4 '17 at 15:27
• Well Wolfram Alpha doesn't really need a good closed form to do calculations? – John Meighan Dec 4 '17 at 15:29
• WA can't do everything mate, and if the input/calculations get too complicated it'll quit on you. – Simply Beautiful Art Dec 4 '17 at 15:31