Math Real Analysis Problem, Riemann Sum Integral?

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SUMMARY

The discussion centers on evaluating the limit of a Riemann sum integral, specifically the expression: limit of n->infinity of n*[1/(n+1)^2 + 1/(n+2)^2 + 1/(n+3)^2 + 1/(2n)^2], which converges to 1/2. Participants clarify that the summation can be expressed as limit of n->infinity of the sum from k=n+1 to 2n of n/(k^2). The solution involves recognizing the Riemann sum structure and manipulating the expression to identify the corresponding function for integration.

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  • Understanding of Riemann sums
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  • Basic knowledge of integral calculus
  • Ability to manipulate summations and series
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  • Learn about convergence of series and limits in calculus
  • Explore techniques for manipulating summations
  • Review examples of definite integrals derived from Riemann sums
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Students studying calculus, particularly those focusing on real analysis and integral calculus, as well as educators looking for detailed explanations of Riemann sums and their applications.

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Part 1. Homework Statement

The problem literally states...

"
The Integral.
limit of n-> infinity of n*[1/(n+1)^2 + 1/(n+2)^2 + 1/(n+3)^2 + 1/(2n)^2] = 1/2
"

According to the teacher, the answer is 1/2. I don't know why or how to get there.


Part 2. The attempt at a solution

I was able to get this far.

limit of n->infinity of the sum from k=n+1 to 2n of n/(k^2)

I put together a summation from whatever info I had, and somehow, that equals 1/2.

I think you are supposed to use Riemann Sum Integral or something like that.

Please explain or show me how with as much detail as possible if you can.

Thank you very much. This is my first post so sorry if it is rude or something is not done right. I read the sticky though! Sorry for the hassles.
 
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[tex]n\cdot\left(\frac{1}{(n+1)^2} + ... + \frac{1}{(2n)^2}\right) = \frac{1}{n}\cdot\left[ \left(\frac{n}{n+1}\right)^2 + ... + \left(\frac{n}{n + n}\right)^2\right][/tex]

This almost obviously looks like a Riemann sum (over what partition?), but you need to make one further manipulation to really see the corresponding function for which we're essentially taking the definite integral of.
 

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