$\sum_{n=1}^\infty\frac{n}{(2n-1)16^n}\binom{2n}{n}^2\left(\sum_{k=n}^\infty\frac{2^k}{k\binom{2k}{k}}\right)=1-\sqrt2+\log(1+\sqrt2).$ 
Prove:$$\sum_{n=1}^\infty\frac{n}{(2n-1)16^n}\binom{2n}{n}^2\left(\sum_{k=n}^\infty\frac{2^k}{k\binom{2k}{k}}\right)=1-\sqrt2+\log(1+\sqrt2).$$

I'm sorry that I don't even know how to start. I haven't met this kind of series before. I've learnt some simplier methods for calculating the value of some simplier series through complex analysis, for example using the Taylor expansion for some holomorphic functions. But it seems that these methods won't work in this problem. So I want to learn more methods or tricks to handle this kind of problems. Would you please recommend some books about this topic?
 A: This is not an answer but it is too long for a comment.
Looking at the lhs, I was expecting a lot of possible things expect what is the rhs.
$$\sum_{k=n}^\infty\frac{2^k}{k \binom{2 k}{k}}=\frac{2^n }{n \binom{2 n}{n}}\,\, _2F_1\left(1,n;n+\frac{1}{2};\frac{1}{2}\right)$$ where appears the gaussian hypergeometric function.
The summand is then
$$a_n=\frac{ \binom{2 n}{n} }{(2n-1)\, 8^n}\,\, _2F_1\left(1,n;n+\frac{1}{2};\frac{1}{2}\right)$$
$$a_n=(-1)^n \frac{ \binom{-\frac{1}{2}}{n}}{2^n\,(2 n-1)}\,\, _2F_1\left(1,n;n+\frac{1}{2};\frac{1}{2}\right)$$
and, for any $n$ (this just comes from the hypergeometric function)
$$a_n=\alpha _n +\beta_n \pi$$ were the $\alpha _n$ and $\beta_n$ are rational numbers. For example,
$$\sum_{n=1}^{10} a_n=\frac{47 (134765953875 \pi -208227582208)}{21646635171840}$$ which is $0.467147$ while the value of the rhs is $0.4671600$.
I do not see how to exploit the fact that
$$\, _2F_1\left(1,n;n+\frac{1}{2};\frac{1}{2}\right)=\sum_{p=1}^{\infty}\frac{ \left(n-\frac{1}{2}\right)! \,(n+p-1)!}{2^p\,(n-1)!
   \left(n+p-\frac{1}{2}\right)!}$$
At this point, I am completely stuck.
A: Not an answer
Are you able to do anything with the following results??
$$ \sum_{n=1}^{\infty}\frac{(-1)^n}{2^{2n}(2n-1)} \binom{2n}{n} = 1-\sqrt{2} $$
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{2n}(2n+1)} \binom{2n}{n} = \ln(1+\sqrt{2}) $$
All that remains is to arrange the terms in a way that would be equal to your identity.
On a Side Note.
Looking at the inner sum :
$$ \sum_{k=1}^{\infty} \frac{2^k}{k \binom{2k}{k}} = \frac{\pi}{2}$$
$$ \sum_{k=2}^{\infty} \frac{2^k}{k \binom{2k}{k}} = \frac{\pi}{2}-1$$
$$ \sum_{k=3}^{\infty} \frac{2^k}{k \binom{2k}{k}} = \frac{\pi}{2}-\frac{4}{3}$$
It looks very similar to examples such as this :
$$ S=\int_{0}^{1} \frac{\ln(x)}{\sqrt{1-x^2}} = -\frac{\pi }{2}\ln(2) $$
Using
$$\int_{0}^{1}\ln(x)\sum_{n=0}^{\infty} \frac{x^{2n}}{2^{2n}}\binom{2n}{n}$$
we end up with nice fractions that end up giving us
$$S = -\sum_{n=0}^{\infty} \frac{1}{2^{2n} (2n+1)^{2}}\binom{2n}{n}$$
BUT if we use the series expansion of the $\ln(x)$
$$ -\int_{0}^{1} \frac{1}{\sqrt{1-x^2}}\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}(x-1)^{n} $$
we have
$$\int_{0}^{1} \frac{(x-1)}{\sqrt{1-x^2}} = 1-\frac{\pi}{2} $$
$$\int_{0}^{1} \frac{(x-1)^{2}}{\sqrt{1-x^2}} = -2+\frac{3\pi}{4} $$
and so on which i find to be similar , so maybe there is something going on there.
Anyways good luck!
