SOLUTION TO PROBLEM H-2

This is a telescoping series (in disguise). For a>1 we have

$$\begin{eqnarray*}\frac{2^n}{a^{2^n}+1} & = & \frac{2^n(a^{2^n}-1)}{a^{2^{n+1}}-1... ... & = & \frac{2^{n}}{a^{2^{n}}-1}-\frac{2^{n+1}}{a^{2^{n+1}}-1} \end{eqnarray*}$$

So

$$S=\sum_{n=0}^\infty \left (\frac{2^{n}}{a^{2^{n}}-1}-\frac{2^{n+1}}{a^{2^{n+1}}-1} \right ) = \frac{1}{a-1}$$

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

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Last modified:   Sat Jan 26 12:43:42 CST 2002