Can please someone help me with this problem: Prove: $\displaystyle \sum_{k=1}^n \dfrac{1}{k} = \sum_{k=1}^n (-1)^{k-1} {n \choose k} \dfrac{1}{k}$
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Originally Posted by barcica Can please someone help me with this problem: Prove: $\displaystyle \sum_{k=1}^n \dfrac{1}{k} = \sum_{k=1}^n (-1)^{k-1} {n \choose k} \dfrac{1}{k}$ Off-hand I'd say there's a typo in this.
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