Sum to infinity

Calculus Level 4

$\large 1 + \dfrac{2^p}{2!} + \dfrac{3^p }{3! }+ \dfrac{4^p}{4!} + \cdots = f(p+1) \cdot e$

The function $f(p+1)$ , as defined above, denotes the $(p+1)$ th term of a well known integer sequence. What is this integer sequence?

For more information, check my "New formula decrypted" note.

None of these. Triangular Numbers Bell Numbers Fibonacci numbers Repunit Numbers

1 solution

Chew-Seong Cheong
Oct 25, 2017

The function $f(p+1)$ gives the $(p+1)$ th Bell number . It is a form of Dobiński's formula which gives the $n$ th Bell number $B_n$ as follows:

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

yes, but i simplified the formula

Sourasish Karmakar - 3 years, 7 months ago

I forgot about solving this problem and I tried the problem e game and solve for $\displaystyle s_n = \sum_{k=0}^\infty \frac {k^n}{k!}$ and prove that $\displaystyle s_n = \sum_{k=0}^{n-1} {n-1 \choose k}s_k$ . I thought I found a new series of number. Then, when I google 1, 1, 2, 5, 52, ... and found that they are Bell number and that $\displaystyle B_n = \sum_{k=0}^{n-1} {n-1 \choose k}B_k$ . Check out my solution there for the proof.

Chew-Seong Cheong - 3 years, 7 months ago

Sum to infinity

For more information, check my "New formula decrypted" note.

1 solution

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