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

$$\int^{1}_{0} (1+x)^n dx=\frac{(1+x)^{n+1}}{n+1}\bigg|^{1}_{0}=\frac{(1+1)^{n+1}-(1+0)^{n+1}}{n+1}=\frac{2^{n+1}-1}{n+1}$$

With indefinite integrals, there is always a constant of integration walking around. Hence, we use definite integrals so that the equality is kept.

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}

Among $n+1$ people, a subset is randomly selected (i.e., each person will be in the subset or not with probability $1/2$). Then one person in the subset (if it is nonempty) is selected at random to win a prize. What's the probability that I (one of the $n+1$ people) win it?

There are $\binom{n}{k}$ ways to pick a subset of size $k+1$ that contains me; the probability of that subset is $\frac{1}{2^{n+1}}$, and the probability that I am the one selected is $\frac{1}{k+1}$. So the desired probability is $$ \frac{1}{2^{n+1}} \sum_{k=0}^n \frac{1}{k+1} \binom{n}{k}. \tag{1} $$ On the other hand, everyone out of the $n+1$ has an equal chance of winning, and there is only a $\frac{1}{2^{n+1}}$ chance of no one being selected, so the probability is $$ \frac{2^{n+1} - 1}{2^{n+1}} \cdot \frac{1}{n+1}. \tag{2} $$ Thus (1) and (2) are equal, and if we multiply by $2^{n+1}$ we get $$ \sum_{k=0}^n \frac{1}{k+1} \binom{n}{k} = \frac{2^{n+1} - 1}{n+1}. $$