What's the value of \sum_{k=1}^\infty\frac{k^2}{k!} ?

Madeline Lott

Madeline Lott

Answered question

2022-01-17

Whats

Answer & Explanation

Raymond Foley

Raymond Foley

Beginner2022-01-17Added 39 answers

The value of Tn=k=1knk! is Bne, where Bn is the n-th Bell number.
To see this, note that
Tn+1=k=1kn+1k!=k=0(k+1)nk!
=k=01k!j=0n(nj)kj
=j=0n(nj)k=1kjk!
=j=0n(nj)Tj
This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by e.
Beverly Smith

Beverly Smith

Beginner2022-01-18Added 42 answers

A basic technique in real (complex) analysis is term by term differentiation of power series:
ez=k=0zkk!
ez=(ez)=k=1kzk1k!
ez=(ez)=k=1k(k1)zk2k!
Evaluating at z=1, one immediately has
e=k=01k!
e=k=1k1k!
e=k=1(k2k)1k!
Combining the second and third equalities, we have the answer.
alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

Just to give a slightly different approach, n=1n2n!=n=1n(n1)! m=0m+1m!=m=0mm!+e m=1mm!+e=m=11(m1)!+e =k=01k!+e=e+e

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