Finding the sum of this alternating series with factorial denominator.

What is the sum of this series?

$$ 1 – \frac{2}{1!} + \frac{3}{2!} – \frac{4}{3!} + \frac{5}{4!} – \frac{6}{5!} + \dots $$

Solutions Collecting From Web of "Finding the sum of this alternating series with factorial denominator."

Hint: We have
$$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots.$$
Multiply both sides by $x$ and differentiate.

Alternatively, write it as:

$$1-\frac{1}{1!} +\frac{1}{2!} – \frac{1}{3!}… +\\
\left(-\frac{1}{1!}+ \frac{2}{2!} – \frac{3}{3!}…\right)$$

The first line is $e^{-1}$ and the second line, after cancelling terms, you see is $-e^{-1}$

More generally, if $$(z)_i = z(z-1)…(z-(i-1))$$ is the falling factorial, and $p(z) = a_0(z)_0 + a_1(a)_1 + … a_k(z)_k$, then:

$$\sum_{n=0}^\infty \frac{p(n)}{n!} x^n =
e^x (a_0 + a_1x + a_2x^2 + … a_k x^k)$$

In this case, $p(z) = 1 + z = (z)_0 + (z)_1$ so

$$\sum_{n=0}^\infty \frac{n+1}{n!} x^n =
e^x (1+x)$$

And, in particular, for $x=-1$, $$\sum_{n=0}^\infty \frac{(-1)^n(n+1)}{n!} = 0$$

Maybe one can do it without using power series:
$$
\begin{align}
\sum_{n=0}^{\infty}(-1)^n\frac{n+1}{n!}
&=\sum_{n=0}^{\infty}(-1)^n\frac{n}{n!}+\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!}\\
&=\sum_{n=1}^{\infty}(-1)^n\frac{1}{(n-1)!}+\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!}\\
&=\sum_{k=0}^{\infty}(-1)^{k+1}\frac{1}{k!}+\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!}\\&=0.
\end{align}
$$