NIMO 2012 A1

$\begin{aligned} \sum_{1<b<a<\infty} \frac 1{2^a5^b} & = \sum_{b=1}^\infty \frac 1{5^b} \sum_{a=b+1}^\infty \frac 1{2^a} \\ & = \sum_{b=1}^\infty \frac 1{5^b} \cdot \frac 1{2^{b+1}} \cdot \frac 1{1-\frac 12} \\ & = \sum_{b=1}^\infty \frac 1{2^b5^b} \\ & = \sum_{b=1}^\infty \frac 1{10^b} \\ & = \frac 1{10} \cdot \frac 1{1-\frac 1{10}} \\ & = \frac 1{10} \cdot \frac {10}9 \\ & = \boxed{\frac 19} \end{aligned}$

$i$ ' $^{th}$ term of the sequence is $\dfrac{1}{2^{i+3}}\left (1-\dfrac {1}{5^i}\right ) =\dfrac{1}{8}\left (\dfrac{1}{2^i}-\dfrac {1}{10^i}\right )$

Therefore the required sum is $\dfrac {1}{8}\left (\dfrac {1}{1-\frac{1}{2}}-\dfrac {1}{1-\frac{1}{10}}\right ) =\boxed {\dfrac{1}{9}}$ .