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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.

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up vote 5 down vote accepted

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


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:


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.

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

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.

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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 $0\le x\le1$: $$ \begin{align} \lim_{n\to\infty}\sum_{k=1}^n\left(\frac{k}{n^2}-\frac{k^2}{n^3}\right) &=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac{k}{n}-\frac{k^2}{n^2}\right)\frac1n\\ &=\int_0^1(x-x^2)\,\mathrm{d}x\\ \end{align} $$

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