Skip to main content
1 of 4
Alapan Das
  • 1.8k
  • 1
  • 7
  • 16

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

$$=\sum_{k=1}^{n} \frac{1}{k}(\sum_{j=1}^{k} \binom{k}{j})$$

$$=\sum_{j=1}^{n} \sum_{k=j}^{n} \binom{k}{j}\frac{1}{k}$$

$$=\sum_{j=1}^{n} \frac{1}{j}(\sum_{k=j}^{n} \binom{k-1}{j-1})$$

$$=\sum_{j=1}^{n} \frac{1}{j} \binom{n}{j}$$

Alapan Das
  • 1.8k
  • 1
  • 7
  • 16