<s>[INST] Evaluating $\sum\limits_{k=0}^\infty n^{2^k}\prod\limits_{m=0}^k\frac1{n^{2^m}+1}$ I found this particular series in a book which asks to evaluate:
$$S=\lim_{k \to \infty}\frac{n}{n+1}+\frac{n^2}{(n+1)(n^2+1)}+\frac{n^4}{(n+1)(n^2+1)(n^4+1)}+...+\frac{n^{2^k}}{(n+1)(n^2+1)...(n^{2^k}+1)}$$
which I believe is equal to 
$${\lim_{k \to \infty}\sum_{j=0}^k\frac{n^{2^j}}{\prod_{m=0}^j(n^{2^m}+1)}}$$
So, I did it like this:
$$\frac{n}{n+1}+\frac{n^2}{(n+1)(n^2+1)}+\frac{n^4}{(n+1)(n^2+1)(n^4+1)}+...+\frac{n^{2^k}}{(n+1)(n^2+1)...(n^{2^k}+1)}$$
$$=\frac{(n+1)-1}{n+1}+\frac{(n^2+1)-1}{(n+1)(n^2+1)}+\frac{(n^4+1)-1}{(n+1)(n^2+1)(n^4+1)}+...+\frac{(n^{2^k}+1)-1}{(n+1)(n^2+1)...(n^{2^k}+1)}$$
Now breaking this gives 
$$1-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{(n+1)(n^2+1)}+\frac{1}{(n+1)(n^2+1)}-\frac{1}{(n+1)(n^2+1)(n^4+1)}...+\frac{1}{(n+1)(n^2+1)...(n^{2^{k-1}}+1)}-\frac{1}{(n+1)(n^2+1)...(n^{2^{k-1}}+1)(n^{2^{k}+1})}$$
which telescopes to 
$$S=1-\frac{1}{(n+1)(n^2+1)...(n^{2^{k-1}}+1)(n^{2^{k}+1})}$$
which gives $$\lim_{k \to \infty}S=1$$
Is this correct? If not, Could somebody help me in the right direction?
 [/INST] Indeed, what you did shows that, for every $n\geqslant1$, $$S=1$$ However, for every $0\leqslant n<1$, $$S=1-\frac1{T(n)}$$ where $$T(n)=\prod_{k=0}^\infty(1+n^{2^k})$$ Now, for every $K$, $$\prod_{k=0}^K(1+n^{2^k})=\sum_{i=0}^{2^{K+1}-1}n^i=\frac{1-n^{2^{K+1}}}{1-n}$$ hence,  for every $0\leqslant n<1$, $$T(n)=\frac1{1-n}$$ and finally, for every nonnegative $n$, $$S=\min\{n,1\}$$
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