Sum Things Are Easier Than They Appear

Algebra Level 3

\[\large \alpha = \sum_{i=1}^\infty \frac 1{2^i} + \sum_{j=1}^\infty \frac 1{2^{j+1}} + \sum_{k=1}^\infty \frac 1{2^{k+2}} + \cdots \]

Let \(\alpha\) as defined above. Find \(\displaystyle \sum_{n=1}^\infty \frac 1{2^{n \alpha}} \).

Give your answer to 3 decimal places.

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