Find sum $ \sum\limits_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2k \rfloor} \cdot 4^{\lfloor \log_2(\log_2k )\rfloor}} $

Let's write $$\sum_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2(k) \rfloor}4^{\lfloor \log_2(\log_2(k))\rfloor}} = \sum_{i=0}^{n-1} \sum_{k=2^{2^i}}^{2^{2^{i+1}}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}4^{\lfloor \log_2(\log_2(k))\rfloor}} + \frac{1}{2^{2^n}4^{n}}$$

$$= \sum_{i=0}^{n-1} \sum_{k=2^{2^i}}^{2^{2^{i+1}}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}4^{i}} + \frac{1}{2^{2^n}4^{n}}$$

Moreover for all $i=0, ..., n-1$, $$\sum_{k=2^{2^i}}^{2^{2^{i+1}}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}} = \sum_{j=2^i}^{2^{i+1}-1} \sum_{k=2^j}^{2^{j+1}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}} = \sum_{j=2^i}^{2^{i+1}-1} \sum_{k=2^j}^{2^{j+1}-1} \frac{1}{2^j} = \sum_{j=2^i}^{2^{i+1}-1} \frac{2^j}{2^j} = 2^i $$

You deduce that $$S = \sum_{i=0}^{n-1} \frac{1}{2^i} + \frac{1}{2^{2^n}4^{n}} = 2 - \frac{1}{2^{n-1}} + \frac{1}{2^{2^n}4^{n}}$$