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?
 A: 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:
$$\sum_{k=0}^{n}\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}=\frac{2^{n+1}-1}{n+1}.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\color{#66f}{\large\sum_{k = 0}^{n}{1 \over k+1}{n \choose k}}&
=\sum_{k = 0}^{n}{n \choose k}\
\overbrace{\pars{\int_{0}^{1}t^{k}\,\dd t}}
^{\ds{=\ \color{#c00000}{1 \over k + 1}}}\ =\ 
\int_{0}^{1}\bracks{\sum_{k = 0}^{n}{n \choose k}t^{k}}\,\dd t
\\[3mm] & =\int_{0}^{1}\pars{1 + t}^{n}\,\dd t =
\left.{\pars{1 + t}^{n + 1} \over n + 1}
\right\vert_{\, t\ =\ 0}^{\, t\ =\ 1}
\\[3mm] & =
{\pars{1 + 1}^{n + 1} - \pars{1 + 0}^{n + 1} \over n + 1}\
=\
\bbox[15px,border:1px solid navy]{\color{#44f}{\large{2^{n + 1} - 1 \over n + 1}}}
\end{align}
A: $$\sum_{k=0}^n\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\sum_{k=0}^n\frac{n+1}{k+1}\binom{n}{k}=$$
$$=\frac{1}{n+1}\sum_{k=0}^n\binom{n+1}{k+1}=\frac{1}{n+1}\left(\sum_{k=1}^{n+1}\binom{n+1}{k}\right)=$$
$$=\frac{1}{n+1}\left(-1+\binom{n+1}{0}+\sum_{k=1}^{n+1}\binom{n+1}{k}\right)=$$
$$=\frac{1}{n+1}\left(-1+\sum_{k=0}^{n+1}\binom{n+1}{k}\right)=$$
$$=\frac{1}{n+1}(-1+2^{n+1})=\frac{2^{n+1}-1}{n+1}$$
