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Hello everyone how can I calculate the sum of:

$\sum_{k=0}^{n}\binom{n}{k}\frac{1}{k+1}$

I tried to use pascal identity and the binomial theorem and didn't success.

Someone can help me please?

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Notice that : $${n \choose k}\frac{1}{k+1}={n+1 \choose k+1}\frac{1}{n+1}$$ The sum is $$\sum_{k=0}^{n}{n+1 \choose k+1}\frac{1}{n+1}=\frac{2^{n+1}-1}{n+1}$$


Another possible solution is $$(1+x)^n=\sum_{k=0}^{n}{n \choose k}x^k$$ We can integrate both sides to get $$\int_0^1(1+x)^ndx=\sum_{k=0}^n{n \choose k}\int_0^1x^kdx=\sum_{k=0}^n{n \choose k}\frac{1}{k+1}$$ The LHS is simply $\frac{2^{n+1}-1}{n+1}$ which gives the final result. Hope this helps.

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