Sum 2

We rewrite $\left(1+a_n\right)$ :

$\begin{aligned} 1+a_n &=1+\left(-1\right)^{n+1}\sum_{k=0}^n {\left(\binom{2n+1}{2k}2^k\right)} \\ &=1+\left(-1\right)^{n+1}\sum_{k=0}^n {\left(\binom{2n+1}{2k}\left(\sqrt{2}\right)^{2k}\right)} \\ &=1+\frac{\left(-1\right)^{n+1}}{2}\sum_{k=0}^n {\left(\binom{2n+1}{2k}\left(\left(\sqrt{2}\right)^{2k}+\left(-\sqrt{2}\right)^{2k}\right)\right)} \\ &=1+\frac{\left(-1\right)^{n+1}}{2}\sum_{k=0}^{2n+1} {\left(\binom{2n+1}{k}\left(\left(\sqrt{2}\right)^{k}+\left(-\sqrt{2}\right)^{k}\right)\right)} \\ &=1+\frac{\left(-1\right)^{n+1}}{2}\left(\sum_{k=0}^{2n+1} {\left(\binom{2n+1}{k}\left(\sqrt{2}\right)^{k}\right)}+\sum_{k=0}^{2n+1} {\left(\binom{2n+1}{k}\left(-\sqrt{2}\right)^{k}\right)}\right) \\ &=1+\frac{\left(-1\right)^{n+1}}{2}\left({\left(1+\sqrt{2}\right)}^{2n+1}+{\left(1-\sqrt{2}\right)}^{2n+1}\right) \\ &=\frac{\left(-1\right)^{n+1}}{2}\left({\left(1+\sqrt{2}\right)}^{2n+1}+{\left(1-\sqrt{2}\right)}^{2n+1}+2{\left(-1\right)}^{n+1}\right) \\ &=\frac{\left(-1\right)^{n+1}}{2}\left({\left(1+\sqrt{2}\right)}^n-{\left(1-\sqrt{2}\right)}^n\right)\left({\left(1+\sqrt{2}\right)}^{n+1}-{\left(1-\sqrt{2}\right)}^{n+1}\right) \\ &=4\left(-1\right)^{n+1}\left(\frac{1}{2\sqrt{2}}\left({\left(1+\sqrt{2}\right)}^n-{\left(1-\sqrt{2}\right)}^n\right)\right)\left(\frac{1}{2\sqrt{2}}\left({\left(1+\sqrt{2}\right)}^{n+1}-{\left(1-\sqrt{2}\right)}^{n+1}\right)\right) \end{aligned}$

So if we define $\left\{b_n\right\}$ as:

$b_n:=\frac{1}{2\sqrt{2}}\left({\left(1+\sqrt{2}\right)}^n-{\left(1-\sqrt{2}\right)}^n\right)$

then, $\left(1+a_n\right)=4\left(-1\right)^{n+1}b_n b_{n+1}$ , and the desired sum is:

$\begin{aligned} S &=\sum_{n=1}^{\infty} \left(\frac{1}{1+a_n}\right) \\ &=\sum_{n=1}^{\infty} \left(\frac{1}{4\left(-1\right)^{n+1}b_n b_{n+1}}\right) \\ &=\frac{1}{4}\sum_{n=1}^{\infty} \left(\frac{{\left(-1\right)}^{n+1}}{b_n b_{n+1}}\right) \end{aligned}$

For the sequence $\left\{b_n\right\}$ , since $r=1\pm\sqrt{2}$ satisfies the characteristic equation $r^2=2r+1$ , we have the recurrence:

$b_n=2b_{n-1}+b_{n-2}$

so the sequence begins $1$ , $2$ , $5$ , $12$ , $29$ , $70$ , $169$ , $408$ , $...$ , which we recognize as denominators of fractions that converge to $\sqrt{2}$ . with the sequence of respective numerators $\left\{c_n\right\}$ beginning $1$ , $3$ , $7$ , $17$ , $41$ , $99$ , $239$ , $577$ , $...$ , which in fact satisfies the same recurrence.

Since the continued fraction expansion of $\sqrt{2}$ is $\left[1;2,2,2,2,2,...\right]$ , the fractions referred to above are the continued fraction convergents to $\sqrt{2}$ , and we therefore have the theorem that $b_{n+1}c_n-b_n c_{n+1}=\left(-1\right)^n$ . Thus:

$\begin{aligned} S &=\frac{1}{4}\sum_{n=1}^{\infty} \left(\frac{{\left(-1\right)}^{n+1}}{b_n b_{n+1}}\right) =-\frac{1}{4}\sum_{n=1}^{\infty} \left(\frac{b_{n+1}c_n-b_n c_{n+1}}{b_n b_{n+1}}\right) =-\frac{1}{4}\sum_{n=1}^{\infty} \left(\frac{c_n}{b_n}-\frac{c_{n+1}}{b_{n+1}}\right) =-\frac{1}{4}\left(\frac{c_1}{b_1}-\frac{c_{\infty}}{b_{\infty}}\right) =\frac{1}{4}\left(\frac{c_{\infty}}{b_{\infty}}-\frac{c_1}{b_1}\right) \\ &=\boxed{\dfrac{1}{4}\left(\sqrt{2}-1\right)} \end{aligned}$

Notice that $4S+1=\sqrt{2}$ , and squaring both sides and simplifying yields:

$\begin{aligned} 16S^2+8S+1&=2 \\ \\ 16S^2+8S&=1 \\ \\ 8S\left(2S+1\right)&=1 \\ \\ S\left(2S+1\right)&=\frac{1}{8} \\ \\ \frac{1}{S\left(2S+1\right)}&=\boxed{8} \end{aligned}$

and the answer is $8$ .