Find the value of $\sum_{k=1}^\infty\frac{k^2}{k!}$

Another way to solve this problem is to exploit the context of power series, particularly term-by-term differentiation.

Observe that: $$ e^z=\sum_{k=0}^{\infty} \frac{z^k}{k!}, \quad e^z=\left(e^z\right)^{\prime}=\sum_{k=1}^{\infty} k \cdot \frac{z^{k-1}}{k!}, \quad e^z=\left(e^z\right)^{\prime \prime}=\sum_{k=1}^{\infty} k(k-1) \frac{z^{k-2}}{k!} . $$

Setting $z=1$, one gets: $$ e=\sum_{k=0}^{\infty} \frac{1}{k!}, \quad e=\sum_{k=1}^{\infty} k \cdot \frac{1}{k!}, \quad e=\sum_{k=1}^{\infty}\left(k^2-k\right) \frac{1}{k!} . $$

Adding the second and third equalities, one has $\sum_{k=1}^{\infty} \frac{k^2}{k!}=2e$.

This answer is, in some ways, "just" a rephrasing of the power series answer by Snoopy, but the broader perspective and name-dropping the relevant tools may be useful.

The relevant tool being generating functions. While not the best book on the topic, Herbert Wilf's generatingfunctionology is at least one readily available source on the topic. There are also a lot of closely related work in combinatorial species and umbral calculus. Generating functions come up absolutely everywhere in many guises and very often let us get entire collections of identities for cheap instead of just individual identities. For most generating functions work, we treat them as formal power series, so the question of their convergence doesn't come up.

So let's actually use it. The $n!$ in the denominator suggests we should consider an exponential generating function. Looking at rules, e.g. the ones listed in generatingfunctionology, we see Rule 2' from that book that states (using standard notation rather than the book's):

If $f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!}$, then $P(xD)f(x) = \sum_{n=0}^\infty P(n) a_n \frac{x^n}{n!}$ where $P$ is a polynomial and $D$ is the differentiation by $x$ operator.

Proving this result is a fairly easy exercise working term-by-term. You can break it down to almost trivial linearity results, the $P(x)Q(x)$ case, and then you simply need to prove that $xDf(x) = \sum_{n=0}^\infty n a_n \frac{x^n}{n!}$ which is easy to verify term-wise.

The proof is straightforward. We note that $e^x$ corresponds to the $a_n\equiv 1$ case, and thus we want $P(n)=n^2$. The above rule immediately gives us that $xDxDe^x$ evaluated at $x=1$ is our desired result. Using that $De^x = e^x$ and the product rule, we easily get $$\begin{align} xDxDe^x &= xDxe^x \\ &= xe^x + x^2De^x \\ &= xe^x + x^2e^x \end{align}$$

Evaluated at $x=1$, this is just $e + e$.

One way to do it is based on the fact that $\displaystyle \sum_{k=0}^\infty \frac1{k!} = e$ and one can do ''change of variable'' in series.

Notice that

\begin{align} \sum_{k=1}^\infty\frac{k^2}{k!} &= \sum_{k=1}^\infty\frac{k}{(k-1)!}&&\text{(cancellation)}\\ &=\sum_{n=0}^\infty\frac{n+1}{n!}&&\text{(change of variable)}\\ &=\sum_{n=0}^\infty\frac{n}{n!}+\sum_{n=0}^\infty\frac{1}{n!}&&\text{(linearity)}\\ &=\left(\sum_{n=1}^\infty\frac{n}{n!}\right)+e \end{align}

Now the problem has been reduced to an ''easier'' one in the sense that the exponent of the numerator goes down to $1$. We can do a similar simplification again:

\begin{align} \sum_{n=1}^\infty\frac{n}{n!} & = \sum_{n=1}^\infty\frac{1}{(n-1)!} = \sum_{n=0}^\infty\frac{1}{k!} = e \end{align}

So $\displaystyle \sum_{k=1}^\infty\frac{k^2}{k!} = e+e=2e$.