3

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?

LM2357
  • 4,083
  • If $n=\frac{1}{2}$ then $S=1-\frac{3}{4}=\frac{1}{4}$ assuming you telescoped right. You need to take care of the case $0\leq n<1$ – kingW3 Jun 11 '17 at 16:51
  • @kingW3 Can you please tell your reasoning behind this?..Maybe the series is defined only for naturals.. – LM2357 Jun 11 '17 at 16:52
  • It could be,that is an easier problem,since for $n=0$ sum is trivially $0$ and for any other integer different from $(-1)$ you have $S=1$ by your formula. – kingW3 Jun 11 '17 at 16:57

2 Answers2

4

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 $n$, $$S=\min\{n,1\}$$

Did
  • 279,727
2

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

kingW3
  • 13,496