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If $n$ is a positive integer, then the above identity holds. I tried to solve this question using generating function.

$$A(x)=\sum_n\left(\sum_{k=1}^n\dfrac{1}{k}\right)x^n=-\dfrac{\log(1-x)}{1-x}$$ $$B(x)=\sum_n\left(\sum_{k=1}^{n}\dfrac{(-1)^{k+1}}{k}{n\choose k}\right)x^n=\log(x+1)(1+x)^n$$

But each side doesn't equal to each other. Did I make any mistake?

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    On the right hand side of the $B(x)$ line, you have an $n$, in the middle you sum over $n$. That doesn't go together well. – Daniel Fischer Jun 01 '14 at 23:43
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    See also http://math.stackexchange.com/questions/809894/help-with-combinatorial-proof-of-identity-sum-k-1n-frac-1k1k –  Jun 02 '14 at 00:07
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    Note that the sum starts from $k=1$. The convolution formula for ordinary generating functions is $\sum_{k=0}^n a_{n-k} b_k$, i.e., the sum starts at $k=0$. – Viktor Vaughn Jun 02 '14 at 00:19
  • Thanks for many helpful info! I solved my question! – Math.StackExchange Jun 02 '14 at 04:07

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