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.

limit as $n$ goes to infinity of $$\sum_{k=1}^n \left(\frac k {n^2}-\frac{k^2}{n^3}\right)$$ I know I need to make this an integral but I cannot figure out how to acquire the limits of integration. Any help would be great.

share|improve this question
add comment

4 Answers 4

up vote 5 down vote accepted

We actually don't need to perform an integral here, we can simply take out constant factors:

$$\lim_{n\to\infty}\left(\sum_{k=1}^{n}\left[\frac{k}{n^{2}}-\frac{k^{2}}{n^{3}}\right]\right)=\lim_{n\to\infty}{\left(\frac{1}{n^{2}}\sum_{k=1}^{n}{k}-\frac{1}{n^{3}}\sum_{k=1}^{n}{k^{2}}\right)}$$

Using our known formulae, we have $\sum_{k=1}^{n}{k}=\frac{n(n+1)}{2}$ and $\sum_{k=1}^{n}{k^{2}}=\frac{n(n+1)(2n+1)}{6}$, so we can rewrite our above expression as:

$$\lim_{n\to\infty}{\left(\frac{n^{2}+n}{2n^{2}}-\frac{n+3n^{2}+2n^{3}}{6n^{3}}\right)}=\lim_{n\to\infty}{\left(\frac{1}{6}-\frac{1}{6n^{2}}\right)}=\frac{1}{6}$$

So we managed to do this particular case without the Riemann Integral. Although I'm sure the other answers will help you to do it using the integral if you are required to do so for homework.

Hope this gives a slightly different and helpful angle on the summation.

share|improve this answer
    
When the Riemann sum is found and evaluated, this makes an excellent verification of the answer. –  robjohn Aug 3 '12 at 23:30
add comment

In evaluating limits of Riemann sums over $[a,b]$, the form of partition you are most likely to come across is the uniform partition, where each interval has length $\frac{b-a}{n}.$ Since $\frac{1}{n}$ is about the most you can factor out of the sum, it seems clear that the sum we should be integrating over $[0,1].$ Now it is your task to find out what the function you're integrating is.

share|improve this answer
add comment

To make this into a Riemann sum, map $x=\frac{k}{n}$ and $\mathrm{d}x=\frac1n$. Then, because $\frac1n\le\frac{k}{n}\le1$, we get $0\le x\le1$: $$ \begin{align} \lim_{n\to\infty}\sum_{k=1}^n\left(\frac{k}{n^2}-\frac{k^2}{n^3}\right) &=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac{k}{n}-\frac{k^2}{n^2}\right)\frac1n\\ &=\int_0^1(x-x^2)\,\mathrm{d}x\\ \end{align} $$

share|improve this answer
add comment

$$\frac1n\sum_{k=1}^n\left[\left(\frac{k}n\right)-\left(\frac{k}n\right)^2\right]$$

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.