I would appreciate if somebody could help me with the following problem
Q:Calculate the sum: $$ \sum_{k=1}^n (-1)^k {n\choose k}\frac{1}{k+1} $$
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2You probably only meant one $(-1)^k$. Also, did you want to sum from $1$ or $0$? – robjohn Dec 12 '13 at 13:24
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2The identity $$\frac{1}{k+1}\binom{n}{k} = \frac{1}{n+1}\binom{n+1}{k+1}$$ could prove helpful – Daniel Fischer Dec 12 '13 at 13:27
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This sum can also be done by using $(k+1)^{-1}= \int_{0}^{1} x^k ~dx$. So the sum=$\int_{0}^{1} (1-x)^n=1/(n+1).$ – Z Ahmed May 10 '19 at 05:34
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Let
$$f(x)=\sum_{k=1}^n (-1)^k {n\choose k}\frac{1}{k+1}x^{k+1}\qquad; f(0)=0$$ so $$f'(x)=\sum_{k=1}^n (-1)^k {n\choose k}x^{k}=(1-x)^n-1$$ hence $$f(x)=\frac{-1}{n+1}(1-x)^{n+1}-x+\frac{1}{n+1}$$
What's $f(1)=-\frac{n}{n+1}$?
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$$ \begin{align} \sum_{k=1}^n(-1)^k\binom{n}{k}\frac1{k+1} &=\sum_{k=1}^n(-1)^k\binom{n+1}{k+1}\frac1{n+1}\\ &=\frac1{n+1}\sum_{k=2}^{n+1}(-1)^{k-1}\binom{n+1}{k}\\ &=\frac1{n+1}\left(1-(n+1)+\sum_{k=0}^{n+1}(-1)^{k-1}\binom{n+1}{k}\right)\\ &=-\frac n{n+1} \end{align} $$
robjohn
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