- #1

Lambda96

- 186

- 64

- Homework Statement
- ##\sum\limits_{k=0}^{n} \frac{(-1)^k}{k+1} \binom{n}{k}=\frac{1}{n+1}##

- Relevant Equations
- none

Hi,

I'm having problems with the proof for the induction of the following problem: ##\sum\limits_{k=0}^{n} \frac{(-1)^k}{k+1} \binom{n}{k}=\frac{1}{n+1}## with ##n \in \mathbb{N}##

I proceeded as follows:

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

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

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

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

Unfortunately, I can't get any further now because I don't know how to solve the term ##\sum\limits_{k=0}^{n+1} \frac{(-1)^k}{k+1} \binom{n}{k-1}## and how to simplify it further. According to mathematica, the term would be ##-\frac{1}{n^2+3n+2}## and if I calculate ##\frac{1}{n+1}-\frac{1}{n^2+3n+2}=\frac{1}{n+2}##, I get the required result.

I'm having problems with the proof for the induction of the following problem: ##\sum\limits_{k=0}^{n} \frac{(-1)^k}{k+1} \binom{n}{k}=\frac{1}{n+1}## with ##n \in \mathbb{N}##

I proceeded as follows:

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

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

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

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

Unfortunately, I can't get any further now because I don't know how to solve the term ##\sum\limits_{k=0}^{n+1} \frac{(-1)^k}{k+1} \binom{n}{k-1}## and how to simplify it further. According to mathematica, the term would be ##-\frac{1}{n^2+3n+2}## and if I calculate ##\frac{1}{n+1}-\frac{1}{n^2+3n+2}=\frac{1}{n+2}##, I get the required result.