Problem H-6 - solution

SOLUTION TO PROBLEM H-6

Let S_n(x) denote the nth partial sum

$\begin{displaymath}S_n(x)=\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)} +\cdots+\frac{x^{2^n}}{(x+1)(x^2+1)\cdots(x^{2^n}+1)}\end{displaymath}$

so that

$\begin{displaymath}\frac{S_n(x)}{x-1}=\frac{x}{x^2-1}+\frac{x^2}{x^4-1} +\cdots+\frac{x^{2^n}}{x^{2^{n+1}}-1}\end{displaymath}$

Using partial fractions

$\begin{displaymath}\frac{x^{2^k}}{x^{2^{k+1}}-1}= \frac{1}{x^{2^{k}}-1}- \frac{1}{x^{2^{k+1}}-1}\end{displaymath}$

So S_n(x)/(x-1) can be seen as a telescoping sum
$\begin{multline*}\frac{S_n(x)}{x-1}= \left (\frac{1}{x-1}-\frac{1}{x^2-1}\right ... ... + \left (\frac{1}{x^{2^{n}}-1}- \frac{1}{x^{2^{n+1}}-1}\right ) \end{multline*}$
We have found that

$\begin{displaymath}\frac{S_n(x)}{x-1} =\frac{1}{x-1}-\frac{1}{x^{2^{n+1}}-1}\end{displaymath}$

Since x>1

$\begin{displaymath}\lim_{n\rightarrow\infty}\frac{S_n(x)}{x-1}=\frac{1}{x-1}\end{displaymath}$

Therefore

$\begin{displaymath}1=\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)}+\frac{x^4}{(x+1)(x^2+1)(x^4+1)} +\cdots\end{displaymath}$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified: Fri Feb 22 10:53:37 CST 2002