Hi I'm currently doing some exercise problems for my first year university math course. The formula I'm supposed to proof is the following:

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

I tried using induction, and while the start is easy, I'm not able to proof the implication from $n$ to $n+1$. Using the pattern we learned for those kinds of questions I'd get

$$ \sum_{k=0}^{n+1} \frac{(-1)^k}{k+1}\binom{n+1}{k} = \sum_{k=1}^n \frac{(-1)^k}{k+1}\binom{n+1}{k} + \frac{(-1)^{n+1}}{n+2}\binom{n+1}{k} $$

The main trouble with that is the $n+1$ in the binomial coefficient where I don't see a way to do induction.

I also thought about multiplying the sum with $n+1$, in the hope that stuff in the sum would cancel out, leaving me with a $1$ at the end, but I didn't progress that way either.

I'd greatly appreciate it if someone could hint me at what I'm missing!


2 Answers 2


Hint: Use $\frac{n+1}{k+1}\binom{n}{k}=\binom{n+1}{k+1}$


Use \begin{eqnarray*} \frac{1}{k+1} = \int_0 ^1 x^k dx. \end{eqnarray*} Now should the sum start at $k=0$ ? ... then we have \begin{eqnarray*} \sum_{k=0}^{n} \frac{(-1)^k}{k+1} \binom{n}{k} &=& \int_0^1 \sum_{k=0}^{n} \binom{n}{k} (-x)^k dx \\ &=& \int_0^1 (1-x)^n dx = \frac{1}{n+1}. \\ \end{eqnarray*}


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