Express $1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$ in a simplifed form I need to express $$1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb  + \frac{1}{n + 1}\binom{n}{n}$$ in a simplified form.
So I used the identity $$(1+x)^n=1 + \binom{n}{1}x + \binom{n}{2}x^2 + \dotsb   + \binom{n}{n}x^n$$
Now on integrating both sides and putting $x=1$.
I am getting $$\frac{2^{n+1}}{n+1}$$ is equal to the given expression.But the answer in my book is  $$\frac{2^{n+1}-1}{n+1}.$$
Where does that -1 term in the numerator come from?

Note that $$\binom{n+1}{k+1}=\frac{n+1}{k+1}\binom{n}{k}$$
Therefore,
\begin{align}
\sum_{k=0}^{n}\frac{1}{k+1}{n\choose k}&=\frac{1}{n+1}\sum_{k=0}^{n}\frac{n+1}{k+1}{n\choose k}
\\&=\frac{1}{n+1}\sum_{k=0}^{n}{{n+1}\choose {k+1}}
\\&=\frac{1}{n+1}\left [\sum_{k=0}^{n+1}{{n+1}\choose k}-1\right ]
\\&=\frac{1}{n+1}\left (2^{n+1}-1\right )
\end{align}
