# Summation calculus: $\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}}$

How can I solve this? $$\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}}$$

Actually I tried many direction, but failed. Please give me some right direction.

$$\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}} = \sum_{k=1}^n \frac{2^{2^{k-1}}}{(1-2^{2^{k-1}})(1+2^{2^{k-1}})}=\cdots$$

## 1 Answer

Let $$S = \sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}} = \sum^{n}_{k=1}\bigg[\frac{(1+2^{2^{k-1}})-1}{1-2^{2^k}}\bigg] = \sum^{n}_{k=1}\bigg[\frac{1}{1-2^{2^{k-1}}}-\frac{1}{1-2^{2^k}}\bigg]$$

which is nothing but Telescopic Sum

So $$S = -1-\frac{1}{1-2^{2^{n}}}$$

• I don't know why I cannot solve it when its values are complex. Thank you so much. – Danny_Kim Apr 15 '17 at 3:23