Is this true: $\ \lim_{n\to\infty} \left( \frac{1}{n} \sum_{i=0}^{n-1} \frac{i}{n-i} \right) = e$?

I was doing the old thing where you play with numbers (with help of my calculator) and came across what seems to be an interesting observation:

$$\ \lim_{n\to\infty} \left( \frac{1}{n} \sum_{i=0}^{n-1} \frac{i}{n-i} \right) = e.$$

At least, this seems to be what is happening with low values of $\ n\ $ with my calculator. So I'm not certain it is correct.

Is this correct, and why? How do we change the left hand side into an integral? I totally forgot how to do this. Or can we prove it with one of the definitions of $\ e\ ?$

Thanks in advance

Solution 1:

This is not true.

Write $S_n = \sum_{i = 0}^{n - 1}\frac i{n - i}$. Adding $1$ to each summand, we get $$n + S_n = \sum_{i = 0}^{n - 1}\frac n{n - i} = n\sum_{i = 1}^n\frac 1 i = nH_n$$where $H_n$ denotes the harmonic number.

From this, it's clear that the limit diverges, as the harmonic numbers do.