Stuck on an alternating sum of binomial coefficients

Question

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!

Answer 1

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

Answer 2

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*}