The value of Tn=∑k=1∞knk! is Bn⋅e, where Bn is the n-th Bell number. To see this, note that Tn+1=∑k=1∞kn+1k!=∑k=0∞(k+1)nk! =∑k=0∞1k!∑j=0n(nj)kj =∑j=0n(nj)∑k=1∞kjk! =∑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.

A basic technique in real (complex) analysis is term by term differentiation of power series:ez=∑k=0∞zkk!ez=(ez)′=∑k=1∞k⋅zk−1k!ez=(ez)″=∑k=1∞k(k−1)zk−2k!Evaluating at z=1, one immediately hase=∑k=0∞1k!e=∑k=1∞k⋅1k!e=∑k=1(k2−k)1k!Combining the second and third equalities, we have the answer.

Just to give a slightly different approach, ∑n=1∞n2n!=∑n=1∞n(n−1)! ∑m=0∞m+1m!=∑m=0∞mm!+e ∑m=1∞mm!+e=∑m=1∞1(m−1)!+e =∑k=0∞1k!+e=e+e

The value of Tn=∑k=1∞knk! is Bn⋅e, where Bn is the n-th Bell number. To see this, note that Tn+1=∑k=1∞kn+1k!=∑k=0∞(k+1)nk! =∑k=0∞1k!∑j=0n(nj)kj =∑j=0n(nj)∑k=1∞kjk! =∑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.

A basic technique in real (complex) analysis is term by term differentiation of power series:ez=∑k=0∞zkk!ez=(ez)′=∑k=1∞k⋅zk−1k!ez=(ez)″=∑k=1∞k(k−1)zk−2k!Evaluating at z=1, one immediately hase=∑k=0∞1k!e=∑k=1∞k⋅1k!e=∑k=1(k2−k)1k!Combining the second and third equalities, we have the answer.

Just to give a slightly different approach, ∑n=1∞n2n!=∑n=1∞n(n−1)! ∑m=0∞m+1m!=∑m=0∞mm!+e ∑m=1∞mm!+e=∑m=1∞1(m−1)!+e =∑k=0∞1k!+e=e+e

Expert Answer to "Whats"

Madeline Lott Answered2022-01-17

Whats

Answer & Explanation

Raymond Foley Expert2022-01-17Added 39 answers

The value of

T_{n} = \sum_{k = 1}^{\infty} \frac{k^{n}}{k!}

B_{n} \cdot e

, where

B_{n}

is the n-th Bell number.
To see this, note that

T_{n + 1} = \sum_{k = 1}^{\infty} \frac{k^{n + 1}}{k!} = \sum_{k = 0}^{\infty} \frac{{(k + 1)}^{n}}{k!}

= \sum_{k = 0}^{\infty} \frac{1}{k!} \sum_{j = 0}^{n} (\begin{matrix} n \\ j \end{matrix}) k^{j}

= \sum_{j = 0}^{n} (\begin{matrix} n \\ j \end{matrix}) \sum_{k = 1}^{\infty} \frac{k^{j}}{k!}

= \sum_{j = 0}^{n} (\begin{matrix} n \\ j \end{matrix}) T_{j}

This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by e.

Beverly Smith

Expert

2022-01-18Added 42 answers

A basic technique in real (complex) analysis is term by term differentiation of power series:

e^{z} = \sum_{k = 0}^{\infty} \frac{z^{k}}{k!}

e^{z} = {(e^{z})}^{'} = \sum_{k = 1}^{\infty} k \cdot \frac{z^{k - 1}}{k!}

e^{z} = (e^{z})^{″} = \sum_{k = 1}^{\infty} k (k - 1) \frac{z^{k - 2}}{k!}

Evaluating at

z = 1

, one immediately has

e = \sum_{k = 0}^{\infty} \frac{1}{k!}

e = \sum_{k = 1}^{\infty} k \cdot \frac{1}{k!}

e = \sum_{k = 1} (k^{2} - k) \frac{1}{k!}

Combining the second and third equalities, we have the answer.

alenahelenash

Expert

2022-01-24Added 366 answers

Just to give a slightly different approach,

\sum_{n = 1}^{\infty} \frac{n^{2}}{n!} = \sum_{n = 1}^{\infty} \frac{n}{(n - 1)!}

\sum_{m = 0}^{\infty} \frac{m + 1}{m!} = \sum_{m = 0}^{\infty} \frac{m}{m!} + e

\sum_{m = 1}^{\infty} \frac{m}{m!} + e = \sum_{m = 1}^{\infty} \frac{1}{(m - 1)!} + e

= \sum_{k = 0}^{\infty} \frac{1}{k!} + e = e + e

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