Prove that $\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}\binom{n}{k} = 1+\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$

Asked Nov 15 '16 at 07:47

Active Nov 15 '16 at 07:47

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For the past 30 minutes I've tried to solve this identity but with no luck. I've tried with math induction,perturbation method and to write the sum and find some simetry but non of them worked. Could someone please help me? Thank you.

$\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}\binom{n}{k} = 1+\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$

asked Nov 15 '16 at 07:47

Dragan Zrilić

https://en.wikipedia.org/wiki/Harmonic_number#Calculation ; http://math.stackexchange.com/questions/1484425/prove-by-induction-sum-limits-k-1n-1k1n-choose-k-frac1k-1-f – Winther Nov 15 '16 at 07:53

Prove that $\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}\binom{n}{k} = 1+\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$

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