Why is $ \lim_{n\to\infty}\sum_{i=0}^{n} \left(\frac i n\right)^{1.5} =\int_{0}^{1} \sqrt{x} \,dx$? Evaluate $$ \lim_{n\to\infty}\sum_{i=0}^{n} \left(\frac i n\right)^{1.5} $$
The solution says this is a Riemann sum for $\int_{0}^{1} \sqrt{x} \,dx$, but I do not understand how they converted the above sum to the definite integral.  
 A: As written, the limit does not exist.  Call that sum $S_n$ for a given value of $n$.  Then $S_{100} \approx 40.5, S_{1000} \approx 400.5$ by direct computation.  One way to see that the limit doesn't exist is as follows:
$$ \sum_{i=0}^n \left( {i\over n} \right)^{1.5} > \sum_{i=n/2}^n \left( {i\over n} \right)^{1.5} > \sum_{i=n/2}^n \left( {n/2 \over n} \right)^{1.5} $$
where the first inequality comes by omitting some positive terms, and the second one since $i\ge n/2$ for all terms in that sum and $i>n/2$ for at least one term.  But the sum is clearly $(n+1) (1/2)^{1.5}$.  So we have $S_n > n/\sqrt{8}$.  
On the other hand each term in the sum is less than or equal to 1 so $S_n \le n+1$.
If you insert a $1/n$ out front you have a Riemann sum and you have
$$\lim_{n \to \infty} {1 \over n} \sum_{i=0}^n \left( {i \over n} \right)^{1.5} = \int_0^1 x^{1.5} dx = 2/5$$
which is probably what the problem is "supposed to" be.  And in fact $S_n \approx 2n/5$.
A: The given limit is diverging. Since $x^{3/2}$ is a Riemann-integrable function over $[0,1]$ we have
$$ \lim_{n\to +\infty}\color{red}{\frac{1}{n}}\sum_{i=0}^{n}\left(\frac{i}{n}\right)^{3/2} = \int_{0}^{1} x^{3/2}\,dx = \frac{2}{5} $$
hence $\sum_{i=0}^{n} i^{3/2} $ behaves like $\frac{2}{5} n^{5/2}$.
