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If $\displaystyle S_{n}=\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)}+\frac{x^{2^{2}}}{(x+1)(x^2+1)(x^{2^2}+1)}+...........+\frac{x^{2^{n}}}{(x+1)(x^2+1)...(x^{2^{n}}+1)}$

Then $\displaystyle \lim_{n\rightarrow \infty}S_{n} = \;,$ Where $x>1$

$\bf{My\; Try::}$ First we will calculate $\bf{r^{th}}$ term of the sequence.

So $$\displaystyle \bf{T_{r}} = \frac{x^{2^{r}}}{(x+1)(x^2+1)............(x^{2^{n}}+1)} = \frac{x^{2^{r}}(x-1)}{x^{2^{r+1}}-1}$$

So We get $$\displaystyle \bf{T_{r}} = \frac{x^{2^{r}}(x-1)}{(x^{2^r}-1)(x^{2^{r}}+1)}$$

Now I did not Understand How can I convert into Telescopic Sum.

Help me

Thanks

Community
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juantheron
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    This may be helpful: $\frac{1}{(x^{2r}-1)(x^{2r}+1)}=\frac{1}{2}\Big(\frac{1}{x^{2r}-1}-\frac{1}{x^{2r}+1}\Big)$. – Albert Dec 21 '15 at 08:22
  • Very interestingly, $$\lim\limits_{n\to\infty},S_n=x$$ for $x\in\mathbb{C}$ such that $|x|<1$. The unit disc $\big{z\in\mathbb{C},\big|,|z|=1\big}$ is the natural boundary, but the series still converges at $x=1$ (to $1$). – Batominovski Oct 10 '19 at 14:57

1 Answers1

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$$\begin{align} S_{\infty}&=\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)}+\frac{x^{2^{2}}}{(x+1)(x^2+1)(x^{2^2}+1)}+\ldots+\frac{x^{2^{n}}}{(x+1)(x^2+1)\cdots(x^{2^{n}}+1)}\\\ &=\frac{x+1-1}{x+1}+\frac{x^2+1-1}{(x+1)(x^2+1)}+\frac{x^{2^{2}}+1-1}{(x+1)(x^2+1)(x^{2^2}+1)}+\ldots+\frac{x^{2^{n}}+1-1}{(x+1)(x^2+1)\cdots}\\\ &=1-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{(x+1)(x^2+1)}+\frac{1}{(x+1)(x^2+1)}-\frac{1}{(x+1)(x^2+1)(x^{2^2}+1)}+\ldots\\\ &=1 \end{align} $$

The last term becomes vanishingly small as we increase $n$, that's why we can ignore it and say that the limit of $S_n$ as $n\to\infty$ is unity.

najayaz
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