$$\lim_{n\to\infty}\sum_{k=1}^n\frac k{k^2+n^2}$$
I need to evaluate this limit, but I don't know how to start. Should i take $1/n^2$ out? Help required. Thank you.
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Clement C.
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How do you take $\frac{1}{n^2}$ out? – Michael Burr Mar 16 '17 at 14:28
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it is a Riemann sum – Dr. Sonnhard Graubner Mar 16 '17 at 14:29
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We haven't studied Riemann sums yet, so how can I evaluate it? – Agnes Van Wood Mar 16 '17 at 14:29
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1@AgnesVanWood The two current answers rely on Riemann sums. If you haven't studied them, please include in your question where this expercise came from (e.g., chapter on XYZ), and what techniques you know/are expected to use (probably this very same XYZ). Otherwise, you are likely to get answers that you don't yet know how to use or reproduce. – Clement C. Mar 16 '17 at 14:34
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@ClementC. Okay, thank you. – Agnes Van Wood Mar 16 '17 at 14:45
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@AgnesVanWood, See also : http://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2/469886#469886 – lab bhattacharjee Mar 16 '17 at 15:11
2 Answers
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Indeed, factor out $1/n^2$. This leaves you with $$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{n}\frac{(k/n)}{1+(k/n)^2}.$$ This resembles a Riemann sum. Under good conditions, we have $$\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{k=1}^n \frac{b-a}{n}f\left(a+k\frac{b-a}{n}\right).$$ From this, we conclude that $a=0$, $b=1$, and $f(x)=\frac{x}{x^2+1}$, so we get $$\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{k^2+n^2}=\int_0^1\frac{x}{1+x^2}\,dx.$$
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Hint: Rewrite $$ \sum_{k=1}^n\frac k{k^2+n^2} = \frac{1}{n}\sum_{k=1}^n\frac{k}{\frac{k^2}{n}+n} = \frac{1}{n}\sum_{k=1}^n\frac{\frac{k}{n}}{\frac{k^2}{n^2}+1} $$ and recognize a Riemann sum for $\int_0^1 f(x)\, dx$, where $f\colon[0,1]\to\mathbb{R}$ is defined by $$ f(x) = \frac{x}{x^2+1}. $$ (This is Riemann integrable on $[0,1]$, since continuous.)
Clement C.
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You're welcome. For the sake of verification, the final limit will be $\int_0^1 f = \frac{\ln 2}{2}$. – Clement C. Mar 16 '17 at 14:35
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1Yes, I've already calculated the final limit and got the same answer. Again, thank you. It was really helpful and incredibly fast answer! – Agnes Van Wood Mar 16 '17 at 14:37