Prove that $ \sum \limits_{n=1}^{\infty} \frac{n}{(n+1)^{(n+2)} (n+2)!}(-1)^{(n+1)} = \frac{23}{24} - \frac 2 3 \sqrt 2 $ Prove that $ \sum \limits_{n=1}^{\infty} \frac{n}{(n+1)^{(n+2)} (n+2)!}(-1)^{(n+1)} = \frac{23}{24} - \frac 2 3 \sqrt 2 $
This question was asked in Math Tripos. and is taken from the classic Hall and Knight  Question Number - 131 of Miscellaneous Examples.
My try:
So, I tried using the expansion of $e^x  = \sum \limits_{i=0}^\infty \frac{x^i}{i!}$ to get factorial part in the denominator and no idea how to get alternate +, - in the series...  My try was based on finding functions which can do that and multiply them to get suitable denominator, but I don't seem to find any. 
Any help is appreciated!
 A: Gathering various informations from the book, I finally figured out the real formula to prove: $$\sum_{n = 1}^\infty \frac{(-1)^{n + 1} (2n - 1)!!}{2^{n + 2}(n + 2)!} = \frac{23}{24} - \frac23\sqrt2,$$ where the $!!$ symbol means double factorial.
For a proof, first rewrite $$(2n - 1)!! = \frac{(2n)!}{(2n)!!} = \frac{(2n)!}{2^nn!}$$ and hence $$\frac{(2n - 1)!!}{2^{n + 2}(n + 2)!} = \frac{(2n)!}{2^{2n + 2}(n + 2)!n!} = \frac{\binom{2n}n}{2^{2n + 2}(n + 1)(n + 2)}.$$
Thus the sum on the left hand side can be written as $$\sum_{n = 1}^\infty (-\frac14)^{n + 1}\frac{\binom{2n}n}{(n + 1)(n + 2)}.$$
The next input is the following famous formula: $$\sum_{n = 0}^\infty \binom{2n}nx^n = (1 - 4x)^{-1/2}.$$ This formula has been discuss many times on this site, see e.g. here.
Integrating the above formula with respect to $x$, we get: $$\sum_{n = 0}^\infty \binom{2n}n\frac{x^{n + 1}}{n + 1} = -\frac 12 ((1 - 4x)^{1/2} - 1).$$
Integrating once more, we get:$$\sum_{n = 0}^\infty \binom{2n}n\frac{x^{n + 2}}{(n + 1)(n + 2)} = \frac1{12}((1 - 4x)^{3/2} - 1) + \frac 12 x.$$
Setting $x = -\frac14$ in the above formula gives the willing result (note the extra term for $n = 0$ on the left).
