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I want to evaluate the following double sum (from blackpenredpen's video) $$\sum_{m=1}^{\infty}{\sum_{n=1}^{m}\frac{1}{2^{m+n}}}$$ In the video, he is showing how to change the order of the sums, but I want to do this normally.

I am pretty confident the inner sum is $\frac{2^m-1}{2^{2m}}$ by geometric series although I am not that good at this subject. Then, how would I evaluate the outer sum?

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Yes indeed we have that

$$\sum_{n=1}^{m}\frac{1}{2^{m+n}}=\frac{1}{2^{m}}\sum_{n=1}^{m}\frac{1}{2^{n}}=\frac{2^m-1}{2^{2m}}$$

and then

$$\sum_{m=1}^{\infty} \frac{2^m-1}{2^{2m}}=\sum_{m=1}^{\infty} \frac{1}{2^{m}}-\sum_{m=1}^{\infty} \frac{1}{2^{2m}}=\sum_{m=1}^{\infty} \frac{1}{2^{m}}-\sum_{m=1}^{\infty} \frac{1}{4^{m}}$$

and we can use geometric series again.

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  • $\begingroup$ oh right, thank you. All the summation "rules" are still new for me. $\endgroup$ – aradarbel10 Nov 30 '19 at 21:56
  • $\begingroup$ @aradarbel10 You did a proper evaluation for the first sum, yu shouldn't have difficulties to obtain the result. You are welcome! Bye $\endgroup$ – user Nov 30 '19 at 21:58

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