Finding $\sum_{k=0}^{\infty} \frac{(-1)^k}{(3+(-1)^k)^k}$

Question

The series $$S=\sum_{k=0}^{\infty} \frac{(-1)^k}{(3+(-1)^k)^k}=1-\frac{1}{2}+\frac{1}{16}-\frac{1}{2^3}+\frac{1}{16^2}-\frac{1}{2^5}+\frac{1}{16^3}-\frac{1}{2^7}+\frac{1}{16^4}-.........$$ can be seen as difference of two infinite convergent geometric series as $$S=\left(1+\frac{1}{16}+\frac{1}{16^2}+\frac{1}{16^3}+\frac{1}{16^4}.....\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^7}+....\right)=\frac{1}{1-\frac{1}{16}}-\frac{1/2}{1-\frac{1}{4}}=\frac{2}{5}.$$ Curioisly the Mathematica doesn't yield this summation. The question: Is this sum all well?

Answer 1

Your approach is fine. Somewhat more elaborated we obtain \begin{align*} \color{blue}{\sum_{k=0}^{\infty}}\color{blue}{\frac{(-1)^k}{\left(3+(-1)^k\right)^k}} &=\lim_{N\to\infty}\sum_{k=0}^{2N-1}\frac{(-1)^k}{\left(3+(-1)^k\right)^k}\\ &=\lim_{N\to\infty}\left(\sum_{k=0}^{N-1}\frac{(-1)^{2k}}{\left(3+(-1)^{2k}\right)^{2k}} +\sum_{k=0}^{N-1}\frac{(-1)^{2k+1}}{\left(3+(-1)^{2k+1}\right)^{2k+1}}\right)\\ &=\lim_{N\to\infty}\left(\sum_{k=0}^{N-1}\frac{1}{4^{2k}} -\sum_{k=0}^{N-1}\frac{1}{2^{2k}}\right)\tag{1}\\ &=\lim_{N\to\infty}\frac{1-\frac{1}{16^N}}{1-\frac{1}{16}} -\frac{1}{2}\lim_{N\to\infty}\frac{1-\frac{1}{4^N}}{1-\frac{1}{4}}\tag{2}\\ &=\frac{1}{1-\frac{1}{16}}-\frac{1}{2}\cdot\frac{1}{1-\frac{1}{4}}\\ &\,\,\color{blue}{=\frac{2}{5}} \end{align*} With (2) the only crucial step where we argue that if both limits exist, then the sum of both limits exist and is equal to the limit (1) of the sum.