<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] What you've done is correct assuming $n\geq 1$ or $n<-1$ when $-1< n <1$ we can write $S$ as $$S=\lim_{k\to\infty}\left(1-\frac{n-1}{(n-1)(n+1)(n^2+1)\cdots(n^{2^k}+1)}\right)\\=\lim_{k\to\infty}\left(1-\frac{n-1}{n^{2^{k+1}}-1}\right)=1-\frac{n-1}{-1}=n$$
</s>