Maybe this is a stupid question, but I don't find my mistake. Here is the situation: I have to calculate a certain probability. The solution is :
$$ \sum_{m=1}^{\infty} \frac{1}{2^m} \sum_{l=km}^{\infty} \frac{1}{2^l} = \sum_{m=1}^{\infty} \frac{1}{2^m} (\sum_{l=0}^{\infty} \frac{1}{2^{l}} - \sum_{l=0}^{km-1} \frac{1}{2^{l}})\sum_{m=1}^{\infty} \frac{1}{2^m}(\frac{1}{\frac{1}{2}}-\frac{1-\frac{1}{2^{km}}}{\frac{1}{2}}) = \sum_{m=1}^{\infty} \frac{1}{2^m} \frac{1}{2^{km-1}} = \frac{2}{2^{k+1}-1} $$
But I have:
$$ \sum_{m=1}^{\infty} \frac{1}{2^m} \sum_{n=m}^{\infty} \frac{1}{2^{kn}} = \sum_{m=1}^{\infty} \frac{1}{2^m} (\sum_{n=0}^{\infty} \frac{1}{2^{kn}} - \sum_{n=0}^{m-1} \frac{1}{2^{kn}})= \sum_{m=1}^{\infty} \frac{1}{2^m}( \frac{1}{1-\frac{1}{2^k}} - \frac{1 - \frac{1}{2^{km}}}{1-\frac{1}{2^k}}) =\sum_{m=1}^{\infty} \frac{1}{2^m}\frac{ \frac{1}{2^{km}}}{1-\frac{1}{2^k}} = \frac{2^k}{(2^k-1)(2^{k+1}-1)}. $$
So could it be that $ \sum_{m=1}^{\infty} \frac{1}{2^m} \sum_{l=km}^{\infty} \frac{1}{2^l} \neq \sum_{m=1}^{\infty} \frac{1}{2^m} \sum_{n=m}^{\infty} \frac{1}{2^{kn}} $ ? or where is my mistake?