What is $\frac{\binom{n}{1}}{1} + \frac{\binom{n}{2}}{2} + \frac{\binom{n}{3}}{3} + \cdots + \frac{\binom{n}{n}}{n}$

Question

Good Day

Today, I learnt that $$\frac{\binom{n}{0}}{1} + \frac{\binom{n}{1}}{2} + \frac{\binom{n}{2}}{3} + \cdots + \frac{\binom{n}{n}}{n + 1} = \frac{2 ^ {n + 1} - 1}{n + 1}$$

I changed it a little and tried to find out $$\frac{\binom{n}{1}}{1} + \frac{\binom{n}{2}}{2} + \frac{\binom{n}{3}}{3} + \cdots + \frac{\binom{n}{n}}{n}$$

Going with a similar approach, I figured out the expression is $$n \cdot {(\frac{\binom{n - 1}{0}}{1} + \frac{\binom{n - 1}{1}}{4}} + \frac{\binom{n - 1}{2}}{9} + \cdots \frac{\binom{n - 1}{n - 1}}{n ^ 2})$$

Similarly, then

1. I integrated $$(1 + x) ^ {n - 1} = \binom{n - 1}{0} + \binom{n - 1}{1}x + \cdots + \binom{n - 1}{n - 1}x ^ {n - 1}$$ from $0$ to $x$.

2. Divide both sides by $x$.

3. Integrate from $0$ to $x$ again.

4. Substitute $x = 1$

But, in the final step, I got stuck on $$\int_{0}^{x}{\frac{(1 + x) ^ {n} - 1}{xn}}dx$$ Putting in WolframAlpha gives a very non-elementary answer.

There are two scenarios.

1. I am wrong in these calculations.
2. This expression does not have a simple value.

I'd appreciate if somebody could help me figure this out. Sorry if I am making a silly mistake, I don't know calculus very well.

Thanks.

Answer 1

As you thought, the answer to your calculation does not have a simple value. What you have donne looks correct to me