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The question is to evaluate $S=\sum_{k=0}^{n} \frac{(-1)^k}{k+1} {n \choose k}$

$\textbf{My Attempt:}$ I have considered the generating function $$ f(x)=\sum_{k=0}^{n} {n\choose k} x^k = (x+1)^n$$Integrated and divided by $x$, $$\frac{1}{x} \int f(x) = \int \frac{{n \choose k}}{k+1} x^k = \left( \frac{1}{x} \right) \frac{(x+1)^{n+1}}{n+1}$$ Finally I considered the value for $x=-1$ which turns out to be $$\frac{1}{-1} \int f(-1) = \left( \frac{1}{-1} \right) \frac{(-1+1)^{n+1}}{n+1}=0$$ Which is wrong. I cannot spot my error

HoopaU
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    @DietrichBurde: That's not a duplicate. The present question asks where the error in the present calculation is, not how to calculate the sum. – joriki Apr 11 '16 at 19:07
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    Yes, this is true. However, I think the answers given at the duplicate explain a lot, and also discuss the integration there - for me still a duplicate in the end. The main question is to evaluate the sum. – Dietrich Burde Apr 11 '16 at 19:09

2 Answers2

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You forgot that indefinite integration is only defined up to an additive constant. You can find that constant by substituting some value at which the function is easy to evaluate. In this case, the sum you want is $0$ at $x=0$, so you actually get

$$ \frac1x\left(\frac{(x+1)^{n-1}-1}{n+1}\right)\;, $$

and then substituting $x=-1$ yields $\frac1{n+1}$ as expected.

Alternatively, you could use the definite integral from $0$ to $1$, which would yield the subtracted term at the lower limit.

joriki
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Try instead $\frac{(-1)^k}{k+1} = \int_{-1}^{0}x^k dx$, then to the summation, then your integral becomes $\int_{-1}^{0}(1+x)^n dx$ and you get the correct result

Alex
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