- #1
steven187
- 176
- 0
Warning:Danger Ahead
hello all
over the last few days Iv been working on this question, but i am seriously left speachless, this will most likely be the hardest question I have ever attempted if anybody is up for a challenge be prepared... alright here we go
we have [tex]\forall n\in\aleph[/tex]
[tex] F(n)= 2n+\frac{2}{3}-e^n\sum_{k=0}^{n-1}\frac{(k-n)^k e^{-k}}{k!}[/tex]
Prove
[tex]\lim_{n\rightarrow\infty} F(n)=0[/tex]
hello all
over the last few days Iv been working on this question, but i am seriously left speachless, this will most likely be the hardest question I have ever attempted if anybody is up for a challenge be prepared... alright here we go
we have [tex]\forall n\in\aleph[/tex]
[tex] F(n)= 2n+\frac{2}{3}-e^n\sum_{k=0}^{n-1}\frac{(k-n)^k e^{-k}}{k!}[/tex]
Prove
[tex]\lim_{n\rightarrow\infty} F(n)=0[/tex]