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Evaluating $ \lim_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $

I recently encountered this question which stumped me because it wasn't obvious what sort series manipulations I could do to find its value. The question is what is the limit as $n$ tends to infinity of $$\sum_{k=0}^{n} \frac{k}{n^2+k^2}$$

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marked as duplicate by no identity, JavaMan, Sasha, DonAntonio, rschwieb Nov 7 '12 at 17:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The body and title ask two different questions. The question in the body makes sense, whereas the title asks about the exact value of a divergent series. –  joriki Nov 7 '12 at 17:04
    
See the respective answer. Just change $N=n^2$ to $N=n$ and $f(x)=n/(n^2+x^2)$ to $f(x)=x/(n^2+x^2)$. Then make some computations. –  no identity Nov 7 '12 at 17:08
    
@Norbert: At least one needs to find the antiderivative of this - different- $f$, so I won't call it a real duplicate. –  Hagen von Eitzen Nov 7 '12 at 17:23

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up vote 3 down vote accepted

$$\lim_{n \to \infty} \sum_{k=0}^n \dfrac{k}{n^2 + k^2} = \lim_{n \to \infty} \sum_{k=0}^n \dfrac{1}{n} \dfrac{\dfrac{k}{n}}{1 + \left(\dfrac{k}{n}\right)^2}.$$

The right side is a Riemann sum: The interval $[0, 1]$ is divided into $n$ subintervals of width $\dfrac{1}{n}$, and rectangles of height $\dfrac{\dfrac{k}{n}}{1 + \left(\dfrac{k}{n}\right)^2}$ are constructed on the subintervals. The sum is the sum of the areas of the rectangles, an approximation to the area under $f(x) = \dfrac{x}{1 + x^2}$ from $x = 0$ to $x = 1$. The limit lets the number of rectangles go to infinity. This gives the integral

$$\int_0^1 \dfrac{x}{1 + x^2}\,dx = \dfrac{1}{2} \ln 2.$$

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