Sum of a Series

All sums will be for $n=0$ to infinity.

Consider the generating function $A(x)=\sum a_nx^n=1+x+\sum a_{n+2}x^{n+2}$ $=1+x+2\sum a_{n+1}x^{n+2}+\sum a_nx^{n+2}-\sum x^{n+2}=2+2x+2x(A(x)-1)+x^2A(x)-\frac{1}{1-x}$ . We solve and find $A(x)=\frac{1-2x}{(1-x)(1-2x-x^2)}$ for $|x|<\sqrt{2}-1$ . For $x=\frac{1}{3}$ we have $S=A\left(\frac{1}{3}\right)=\boxed{2.25}$

An explicit equation for the sequence will be of the form $a_n = k + c_1p_1^n + c_2p_2^n,$ where $k = 2k + k - 1$ and $p_{1,2}$ are the solutions of $p^2 - 2p - 1 = 0$ .

This gives us $k = \tfrac12$ , and $p_{1,2} = 1\pm\sqrt 2$ ; from $a_0 = 1$ we conclude that $c_1 + c_2 = \tfrac12$ ; from $a_1 = 1$ it follows that $c_1 = c_2 = \tfrac14$ . Thus $a_n = \tfrac12 + \tfrac14(1 + \sqrt 2)^n + \tfrac14(1 - \sqrt 2)^n.$ For the infinite sum we note that $\sum_{k=0}^\infty x^k = 1/(1-x)$ , so that $\sum_{k=0}^\infty \frac{1/2}{3^k} = \frac {1/2}{ (1-1/3)} = \tfrac 34; \\ \sum_{k=0}^\infty \frac{\tfrac14 (1 \pm \sqrt 2)^k}{3^k} = \frac{1/4}{1 - (\tfrac13 \pm \tfrac13\sqrt 2)} \\ = \frac{3}{4 \cdot (2 \mp \sqrt 2)} = \frac{3(2 \pm \sqrt 2)}{4\cdot 2} = \tfrac34 \pm \tfrac38\sqrt 2;$ summing these, we find $\sum_{k=0}^\infty a_n = \tfrac 34 + \tfrac34 + \tfrac38\sqrt 2 + \tfrac 34 - \tfrac38\sqrt 2 = \tfrac 94 \approx \boxed{2.25}.$