Question

`\rm \sum_(n = 0)^ \infty \frac{( - 1 {)}^(n) (n + 1)}{(n + 2)n! \cdot {2}^(n - 1) } \\`​

Answer 1

Expand into partial fractions,

`(n+1)/(n+2) = ((n+2)-1)/(n+2) = 1 - \frac1{n+2}`

and rewrite the sum as

`\displaystyle 2 \left(\sum_(n=0)^\infty \frac1{n!} \left(-\frac12\right)^n - \sum_(n=0)^\infty \frac1{n!(n+2)} \left(-\frac12\right)^n\right)`

The first sum we recognize as the series expansion of
`e^x` at
`x=-\frac12`. For the other sum, let

`\displaystyle f(x) = \sum_(n=0)^\infty \frac1{n!(n+2)} x^n`

Introduce factors of
`x` so that differentiating both sides will eliminate the
`n+2` in the denominator of the summand.

`\displaystyle x^2 f(x) = \sum_(n=0)^\infty \frac1{n!(n+2)} x^(n+2) \\\\ ~~~~ \implies x^2 f'(x) + 2x f(x) = \sum_(n=0)^\infty \frac1{n!} x^(n+1) = xe^x`

Solve the differentiating equation for
`f`, using the initial condition
`f(0)=0`.

`\left(x^2 f(x)\right)' = xe^x \implies x^2 f(x) = (x-1)e^x + C`

`f(0) = 0 \implies 0 = -1 + C \implies C = 1`

`\implies f(x) = (x-1)/(x^2) e^x + \frac1{x^2}`

Now let
`x=-\frac12`. We find

`\displaystyle 2 \left(\sum_(n=0)^\infty \frac1{n!} \left(-\frac12\right)^n - \sum_(n=0)^\infty \frac1{n!(n+2)} \left(-\frac12\right)^n\right) \\\\ ~~~~~~~~ = 2 \left(e^(-1/2) - \left(4-6e^(-1/2)\right)\right) = \boxed{(14)/(\sqrt e) - 8}`