# 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?

#### Solutions Collecting From Web of "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 $\Large {{n+1}\choose {k+1}}=\frac{(n+1)!}{(n-k)!(k+1)!}$

$\Large =\frac{n+1}{k+1}\frac{n!}{(n-k)!k!}=\frac{n+1}{k+1}{n\choose k}$, $\quad$ which holds for all positive integers $k\leq n$.

Let $S$ be the sum of the given series.

Then, $S=\displaystyle\sum_{k=0}^{n}\frac{1}{k+1}{n\choose k}=\frac{1}{n+1}\displaystyle\sum_{k=0}^{n}\frac{n+1}{k+1}{n\choose k}$

$=\frac{1}{n+1}\displaystyle\sum_{k=0}^{n}{{n+1}\choose {k+1}}$

$=\frac{1}{n+1}\left [\displaystyle\sum_{k=0}^{n+1}{{n+1}\choose k}-1\right ]$

$=\frac{1}{n+1}\left (2^{n+1}-1\right )$

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
$$\sum_{k=0}^n \frac{1}{k+1} \binom{n}{k} = \frac{2^{n+1} – 1}{n+1}.$$

Remember when you integrate both sides, there will be the constant of integration. You need to first find this constant of integration. That’s where the $-1$ comes from. In actual fact, it is $\frac{-1}{n+1}$ but looks like $-1$ because it’s been absorbed into the numerator.

You may like to try $x=0$ to find this $+C$ value…