Using binomial theorem to evaluate summation $\sum_{k=0}^n \frac{1}{k+1} \binom nk$ in closed form

A problem I'm trying to figure out asks that I use the binomial theorem (or any other method I want) to evaluate $\sum_{k=0}^n \frac{1}{k+1} {n \choose k}$ in closed form.

The binomial theorem stated in my textbook: $(1+x)^n = {n\choose 0}+{n\choose 1}x+{n\choose 2}x^2+...+{n\choose k}x^k+...+{n\choose n}x^n$

I think the idea behind evaluating the summation to closed form is to perform some manipulation on the binomial theorem until it basically looks like what I want. My textbook only gives simplistic examples, like showing that $\sum_{k=0}^n {n \choose k}$ by setting $x=1$. This doesn't really give me a good sense of how to do this problem.

Could I please have a hint?

• Integrate, and see what happens. – Lucian Sep 21 '14 at 17:05
• OK, this has gotten me really close to the answer's below. I'm trying to remember how integration works (haven't done any math in a long time), I'm trying to figure out why the constant from integrating = -1 – GrinReaper Sep 21 '14 at 17:16
• – Martin Sleziak Jan 15 '17 at 9:13

Lucian's comment is perfectly fine. As an alternative, a nice old trick: $$\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\binom{n+1}{k+1}$$ hence: $$\sum_{k=0}^{n}\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}=\frac{2^{n+1}-1}{n+1}.$$
$$\sum_{k=0}^n\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\sum_{k=0}^n\frac{n+1}{k+1}\binom{n}{k}=$$ $$=\frac{1}{n+1}\sum_{k=0}^n\binom{n+1}{k+1}=\frac{1}{n+1}\left(\sum_{k=1}^{n+1}\binom{n+1}{k}\right)=$$ $$=\frac{1}{n+1}\left(-1+\binom{n+1}{0}+\sum_{k=1}^{n+1}\binom{n+1}{k}\right)=$$ $$=\frac{1}{n+1}\left(-1+\sum_{k=0}^{n+1}\binom{n+1}{k}\right)=$$ $$=\frac{1}{n+1}(-1+2^{n+1})=\frac{2^{n+1}-1}{n+1}$$
• The integration in $\left(\,0,1\,\right)$ let's to introduce a well known result. Namely,$\sum_{k = 0}^{n}{n \choose k}t^{k} = \left(\,1 + t\,\right)^{n}$. In many situations, it's easier to reduce the series to an integral. I saw $\color{#00f}{\tt @Sherlock Holmes}$ already pointed out this fact. – Felix Marin Sep 23 '14 at 16:47