Find the values of $\sum_{k=0}^{n} \frac{{n\choose k }}{k+1}$

Question

Let m be a positive integer.Find the values of $$\sum_{k=0}^n \frac{{n\choose k }}{k+1}$$. Leave your answer in terms of n where appropriate.

Remark. There is an alternative method for computing the sums described here: make use of integration.

I can only list out the terms $$\sum_{k=0}^n \frac{{n\choose k }}{k+1}=1+\frac{\binom{n}{1}}{2}+\frac{\binom{n}{2}}{3}+...+\frac{1}{m+1}$$ I can't think of how to simplify them and get the answer.

Also, the question said I can use integration to solve it, but I have no idea how to start.I would greatly appreciate it if someone could show how to solve this.

Answer 1

$$\sum_{k=0}^{n}\frac{{n\choose k}}{k+1}=\frac{1}{n+1}\sum_{k=0}^{n}{n+1\choose k+1}=\frac{1}{n+1}\sum_{j=1}^{n+1} {n+1 \choose j}=\frac{2^{n+1}-1}{n+1}.$$

Answer 2

Hint: You can rewrite the $\frac{1}{1+k}$ factor using the integral

$$\int_0^1 x^k \,dx= \frac{1}{1+k}.$$

Then pull the summation inside the integral.

Answer 3

$$\sum_{k=0}^{n}{n \choose k} x^k= (1+x)^n$$ $$\implies \int_{0}^{1}\sum_{k=0}^{n} {n \choose k} x^k dx=\int_{0}^{1} (1+x)^n dx.$$ $$\implies \sum_{k=0}^{n} \frac{{n \choose k}}{k+1}=\frac{2^{n+1}-1}{n+1}.$$