E(X) = Σ (from x=1 to ∞) (xλ^(x)e^(-λ))/x! = ∑ (x/x!)(λ^(x)e^(-λ)). Since x/x! = 1/(x-1)!, we can write
= ∑ (1/(x-1)!)(λ^(x)e^-λ)
= e^(-λ) ∑ (λ^x/(x-1)!), we can move e^(-λ) across because it is not affected by the index. Next is a little tricky but follow along...
= e^-λ ∑ (from x=0 to ∞) (λ^x+1)/(x-1+1)!
= e^-λ ∑ (λ^x)(λ)/x!
= λe^-λ ∑ λ^x/x!, now we can use the Taylor Series Expansion of eλ, if you have forgotten what it looks like, you can find the solution in any calculus book, including the Stewart Calculus book if thatís what you have. Since eλ = ∑ λ^x/x!, we can state
= λ. ■
Sorry I don't know how to code. If you have any questions on anything just let me know. I hope that it makes sense. If you want to code what you understand the solution to be, I can look it over.