While attempting to prove the Mean Value Theorem, I stumbled across the following limit and associated sum:
$$\lim_{n\to\infty}\sum_{k=0}^n \frac{f(k/n)}{(n+1)}$$
Now, I have solved many summations before of the form
$$\lim_{n\to\infty}\sum_{k=1}^n \frac{f(k/n)}{n} = \int_0^1f(x)\,\mathrm{d}x$$
However, I am unable to put the first summation into a form where I can turn it into a Riemann Integral. As far as I can tell (and through the aid of Mathematica) the first sum with limit seems to be equivalent to the second.... Is there a proof of why this is so?
Note that I have checked $\lim_{n\to\infty}\sum_{k=0}^n \frac{(k/n)^2}{(n+1)} = \frac 13 = \int_0^1 x^2 \mathrm{d}x$ as well as $\lim_{n\to\infty}\sum_{k=0}^n \frac{e^(k/n)}{(n+1)} = e-1 = \int_0^1 e^x \mathrm{d}x$ by way of Mathematica
Hint: $$\sum_{k=0}^n \frac{f(k/n)}{(n+1)}=\frac{f(0)}{n+1}+\sum_{k=1}^{n} \frac{f(k/n)}{n+1}= \frac{f(0)}{n+1}+\frac{n}{n+1}\sum_{k=1}^{n} \frac{f(k/n)}{n}.$$