Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $<\infty$?

share|improve this question
add comment

5 Answers

up vote 2 down vote accepted

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

share|improve this answer
add comment

We have that $$\lim_{a\to\infty}\left(\frac{1}{a}\sum_{b=0}^a\bigg(\frac{b}{a}\bigg)^2\right)=\lim_{a\to\infty}\left(\frac{1}{a^3}\sum_{b=0}^ab^2\right)=\lim_{a\to\infty}\left(\frac{1}{a^3}\cdot\frac{2a^3+3a^2+a}{6}\right)=\lim_{a\to\infty}\left(\frac{1}{3}+\frac{1}{2a}+\frac{1}{6a^2}\right)=\frac{1}{3}$$

share|improve this answer
add comment

Hint: Use Riemann sum of an integral of the function $f(x)=x^2$ on the interval $[0,1]$.

share|improve this answer
add comment

Let us change a little the symbols used to something a little more conventional:

$$\frac{1}{n}\sum_{k=0}^n\left(\frac{k}{n}\right)^2\xrightarrow [n\to\infty]{}\int_0^1x^2\,dx$$

share|improve this answer
add comment

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?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.