For some series, it is easy to say whether it is convergent or not by the "convergence test", e.g., ratio test. However, it is nontrivial to calculate the value of the sum when the series converges. The question is motivated from the simple exercise to determining whether the series $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$ is convergent. One may immediately get that it is convergent by the ratio test. So here is my question:
What's the value of $$\sum_{k=1}^{\infty}\frac{k^2}{k!}?$$
On
Hint 1: $k^2=k(k-1)+k$. Hint 2: simplify the fractions $k(k-1)/k!$ and $k/k!$. Hint 3: write the series expansion around $x=0$ of the function $x\mapsto\exp(x)$.
On
Wolfram Alpha says it is $2e$. Another derivation is to start with $e^x=\sum \limits_{n=0}^\infty \frac{x^n}{n!}$, apply $\frac{d}{dx}x\frac{d}{dx}$ to both sides, and evaluate at $x=1$.
On
The value of $T_n := \displaystyle\sum_{k=1}^{\infty} \frac{k^n}{k!}$ is $B_n \cdot e$, where $B_n$ is the $n^{th}$ Bell number.
To see this, note that
$$\begin{align} T_{n+1} = \sum_{k=1}^{\infty} \frac{k^{n+1}}{k!} &= \sum_{k=0}^{\infty} \frac{(k+1)^n}{k!} \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{j=0}^n {n \choose j} k^j \\ &= \sum_{j=0}^n {n \choose j} \sum_{k=1}^{\infty} \frac{k^j}{k!} \\ &= \sum_{j=0}^{n} {n \choose j} T_j \end{align}$$
This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by $e$.
Edit: I was unaware at the time I posted my answer, but the argument I gave goes by the name of Dobiński's formula.
The sum is equal to $2e$. First of all, the term $k^2/k!$ may be partly canceled as $k/(k-1)!$. Second, this can be written as $(k-1+1)/(k-1)!$. The term $k-1+1$ is divided to two terms. In the first term, the $k-1$ may be canceled again, giving us $e$. The second term leads to $e$ immediately. So the total sum is $2\exp(1)$.
In a similar way, one may easily calculate the sum even if $k^2$ is replaced by $k^n$, any positive integer power of $k$. The result is always a multiple of $e$.