Suppose $\left\{a_n\right\}$ is a sequence defined by:
$$a_n:=\left(-1\right)^{n+1} \sum_{k=0}^n {\left(\binom{2n+1}{2k}2^k\right)}$$
The problem is to evaluate:
$$\sum_{n=1}^{\infty} {\left(\frac{1}{1+a_n}\right)}$$
An estimate for the value of the sum is approximately $0.1036$.
One curious thing I was able to find is that the denominator of the sum can be factored:
$$\left(1+a_n\right)=\frac{1}{2}\left(-1\right)^{n+1}\left({\left(1+\sqrt{2}\right)}^n-{\left(1-\sqrt{2}\right)}^n\right)\left({\left(1+\sqrt{2}\right)}^{n+1}-{\left(1-\sqrt{2}\right)}^{n+1}\right)$$
But I do not know how to use this factorization to solve the sum.
Any ideas?? Thanks!
