Exact Value of Infinite Sum $ \sum_{n = 0}^{\infty } \frac{{(-1)}^n}{n!(n+2)}$

Question

Find exact value of the sum:

$ \sum_{n = 0}^{\infty } \frac{{(-1)}^n}{n!(n+2)} $

We could manipulate as follows:

$ \sum_{n = 0}^{\infty } \frac{{(-1)}^n}{n!(n+2)} = \sum_{n = 0}^{\infty } \frac{{(-1)}^n(n+1)}{(n+2)!} = \sum_{n = 0}^{\infty } \frac{{(-1)}^n(n)}{(n+2)!} + \sum_{n = 0}^{\infty } \frac{{(-1)}^n}{(n+2)!}$

The second term can be computed by integrating the Maclaurin series for $e^x$ twice:

$ \int (\int e^{x}dx) dx = \sum_{n = 0}^{\infty } \frac{{x}^{(n+2)}}{{(n+2)}!}$

which we can rewrite as follows if we set $x=-1$:

$e^{-1}= \sum_{n = 0}^{\infty } \frac{{(-1)}^{n}}{{(n+2)}!}$

As for the first term $\sum_{n = 0}^{\infty } \frac{{(-1)}^n(n)}{(n+2)!}$, any hints would be greatly appreciated, or perhaps if general direction is already incorrect in the first place?

Answer 1

You have

$$\sum_{n = 0}^{\infty } \frac{{(-1)}^n}{n!(n+2)} = \int_0^1xe^{-x}dx = \frac{e-2}{e}$$

Answer 2

Hint: Use $$ \frac{1}{{n!(n + 2)}} = \frac{{n + 1}}{{(n + 2)!}} = \frac{{n + 2 - 1}}{{(n + 2)!}} = \frac{1}{{(n + 1)!}} - \frac{1}{{(n + 2)!}}. $$ instead.

Answer 3

If you rewrite $$ \frac{(-1)^k}{k+2} = \frac{(-1)^{k+2}}{k+2} =\int_{0}^{-1}x^{k+1}dx $$ Interchange integral and sum, you get an expression, solve the sum for $e^x$ $$ \int_{0}^{1}xe^xdx $$