Infinite sum and integral I have a question regarding a sum.
Is the following expression finite and can be calculated?
$$\lim_{a\to\infty}\frac{1}{a}\sum_{b=0}^a \left(\frac{b}{a}\right)^2$$
Could I also approximate the sum by an integral since the upper index grows to infinity like
$$\lim_{a\to\infty}\int_{b=0}^a~db \frac{b^2}{a^3}$$ which would be $<\infty$?
 A: We have that
$$\lim_{a\to\infty}\left(\frac{1}{a}\sum_{b=0}^a\bigg(\frac{b}{a}\bigg)^2\right)=\lim_{a\to\infty}\left(\frac{1}{a^3}\sum_{b=0}^ab^2\right)=\lim_{a\to\infty}\left(\frac{1}{a^3}\cdot\frac{2a^3+3a^2+a}{6}\right)=\lim_{a\to\infty}\left(\frac{1}{3}+\frac{1}{2a}+\frac{1}{6a^2}\right)=\frac{1}{3}$$
A: Hint: Use Riemann sum of an integral of the function $f(x)=x^2$ on the interval $[0,1]$.
A: The Riemann Sum for $f(x)=x^2$ on the interval $[0,1]$ is $$\lim_{N\to\infty}\sum_{i=0}^Nf\left(x_i\right)\cdot\frac{1}{N}=\lim_{N\to\infty}\frac{1}{N}\sum_{i=0}^N\left(\frac{i}{N}\right)^2.$$ Do you notice any similarities? Do you see how to calculate it from here?
A: Let us change a little the symbols used to something a little more conventional:
$$\frac{1}{n}\sum_{k=0}^n\left(\frac{k}{n}\right)^2\xrightarrow [n\to\infty]{}\int_0^1x^2\,dx$$
A: You want the limit
$$\lim_{n\to +\infty}\sum_{k=0}^{n} \frac{k^2}{n^3}$$
When you have sums of the form $\sum_{k=0}^nf(k,n)$ and want to test convergence or even compute them a good idea is to use the definition of Riemann Integrals. Choose your favorite partition of the simplest set you can imagine, mine happens to be
\begin{equation}\mathcal{P}=\left\{ 0=x_0<x_1<...<\frac{i}{n}<...<x_n=1 \right\}\end{equation}
of $[0,1]$. Now we must choose our function $f:[0,1]\to \mathbb{R}$ wisely so that
 $$U_{f,\mathcal{P}}=\sum_{k=1}^n\frac{k^2}{n^3}$$
But
\begin{equation}U_{f,\mathcal{P}}=\sum\limits_{i=1}^{n}{M_i(f)\left( x_i-x_{i-1} \right)}=\sum\limits_{i=1}^{n}\frac{\sup_{x\in [x_{{i-1}},x_i]}f(x)}n
\end{equation}
If we choose an increasing function this simplifies to
$$\sum_{k=1}^n\frac{k^2}{n^3}=\sum_{k=1}^{n}\frac{f(x_k)}n$$
Matching the terms gives
$$f(x_k)=\frac{k^2}{n^2}$$
Now because $x_k=\frac{k}{n}$ it makes sense to substitute $k=nx$ and get the formula for $f$. Doing that yields
$$ f(x)=\frac{x^2n^2}{n^2}=x^2$$
That's how you can come up with $f$. The rest can be found in the other answers
