Calculate the limit using definite integrals: $\lim_{n\to\infty}2n\sum_{k=1}^n\frac1{(n+2k)^2}$

Well, I started like this:

$\lim_{n\to\infty}2n\sum_{k=1}^n\frac1{(n+2k)^2}=\lim_{n\to\infty}2n[\frac1{(n+2)^2}+...+\frac1{(n+2n)^2}]$ From this point and on i'm always getting stuck. I know i should use Riemann integral definition but i don't know how can i recognize which function am i looking at? or how can i recognize the correct $\Delta x$? or at least the "borders" of the definite integral?

Say i presume $\Delta x=\frac1n$ (It's incorrect, Should be $\frac 2n$ - But why?), Then: $\lim_{n\to\infty}\frac1n[\frac{2n^2}{(n+2)^2}+...+\frac{2n^2}{(n+2n)^2}] $, Using limit on both of the 'edges' of the sum should give me the 'borders?' of the definite integral?

As you can see i'm pretty much lost in this kind of questions, BTW: An explanation will be of much more use than only an answer to this one.

  • 1
    $\begingroup$ Your choice of $\Delta x$ as $1/n$ or $2/n$ won't matter. You can factor out the $2$ and just use $1/n$. At least for me, it's conceptually easier to just always use $1/n$. $\endgroup$
    – Potato
    May 25 '13 at 17:48

Factor out an $n^2$ from the denominator of the summand. You then get

$$\frac{2}{n} \sum_{k=1}^n \frac{1}{(1+2 k/n)^2}$$

This is the form of a Riemann sum; take the limit as $n \to \infty$ and identify the result as a definite integral over a variable $x$.

A Riemann sum takes the form

$$\int_a^b f(x) \, dx = \lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f\left(a + (b-a)\frac{k}{n}\right)$$

What is $a$? $b$?

  • $\begingroup$ I guess a and b would be 1 and 3. $\endgroup$ May 25 '13 at 18:10
  • 1
    $\begingroup$ @TheAlchemist: that's one way to look at it ($f(x)=1/x^2$). $\endgroup$
    – Ron Gordon
    May 25 '13 at 18:12

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