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It's e-asy (Posted on 2008-02-19) Difficulty: 3 of 5
Outline a method for calculating Sk = Σ nk/n! for n=1 to infinity and compute the first few terms of Sk for natural numbers k.

See The Solution Submitted by FrankM    
Rating: 4.0000 (2 votes)

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Solution Solution | Comment 4 of 7 |
S(k)=Σ n^k/n! = Σ n^(k-1)/(n-1)! = Σ(n-1+1)^(k-1)/(n-1)!
Now use binomial theorem to expand it.
S(k)=ΣΣC(k-1,i)*(n-1)^i/(n-1)! (i: 0 to k-1)
S(k)=ΣC(k-1,i)Σ(n-1)^i/(n-1)!  {Here I reversed summation over variabbles}
S(k)=ΣC(k-1,i)S(i) ---- (1)
Here the method is to compute the starting terms S(1), S(0) and compute
remaining terms using S(1) recursively.
S(1) =Σn/n! = Σ1/(n-1)! = e,S(0) = e-1.
S(2) = C(1,0) S(0)+ C(1,1) S(1) = e-1+e=2e-1
S(3)=C(2,0) S(0)+C(2,1) S(1)+C(2,2) S(2)=e-1+2e+2e-1=5e-2
and so on.

  Posted by Praneeth on 2008-02-20 01:22:55
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