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That

$$\lim_{n \to \infty}\sum_{k=1}^n \frac{n}{n^2 + k^2} = \frac{\pi}{4},$$

is easily shown by identifying this as the limit of a Riemann sum for $\displaystyle \int_0^1 (1 + x^2)^{-1} \, dx$.

We can find the limit in terms of the integral by bounding and aplying the squeeze principle:

$$\int_{k/n}^{(k+1)/n} \frac{dx}{1+x^2} \leqslant \frac{n}{n^2 + k^2}=\frac{1/n}{1 + (k/n)^2}\leqslant \int_{(k-1)/n}^{k/n} \frac{dx}{1+x^2},$$

leading to

$$\int_{1/n}^{1} \frac{dx}{1+x^2} + \frac{1}{2n}\leqslant \sum_{k=1}^n\frac{n}{n^2 + k^2}\leqslant \int_{0}^{1} \frac{dx}{1+x^2}.$$

Can anyone suggest an "elementary" way to find this limit without using the integral.

Thanks.

RRL
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  • Did you have a look at the references here ? : http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-n-1-infty-frac1n2 – Theo Douvropoulos Apr 06 '15 at 21:34
  • @Theo: Thanks. I'm taking a look. Perhaps what sets this apart is the dependence on $n$ both in the summand and as the summation limit: $\sum_{k=1}^n a_{n,k}$ – RRL Apr 06 '15 at 21:40
  • Yes, I wasn't really saying this is duplicate. More that you can get some nice insight in a similar question -that has received quite a lot of attention... – Theo Douvropoulos Apr 06 '15 at 21:42

2 Answers2

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Using equation $(7)$ from this answer, $$ \begin{align} \sum_{k=1}^\infty\frac{n}{n^2+k^2} &=\frac i2\sum_{k=1}^\infty\left(\frac1{-k+in}+\frac1{k+in}\right)\\ &=-\frac1{2n}+\frac i2\sum_{k\in\mathbb{Z}}\frac1{k+in}\\ &=-\frac1{2n}+\frac i2\pi\cot(\pi in)\\ &=-\frac1{2n}+\frac\pi2\coth(\pi n)\tag{1} \end{align} $$ Then $$ \begin{align} \lim_{n\to\infty}\sum_{k=1}^\infty\frac{n}{n^2+k^2} &=\lim_{n\to\infty}\left(-\frac1{2n}+\frac\pi2\coth(\pi n)\right)\\ &=\frac\pi2\tag{2} \end{align} $$

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robjohn
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Hint: first by Taylor, then by Faulhaber we get the Gregory Series (without rigor): $$\begin{align}\frac1n\sum_{k=1}^n\left(1+\left(\frac kn\right)^2\right)^{-1} &\to\frac1n\sum_{k=1}^n\sum_{j=0}^\infty(-1)^j\left(\frac kn\right)^{2j}\\ &\to\sum_{j=0}^\infty\frac{(-1)^j}{n^{2j+1}}\sum_{k=1}^nk^{2j}\\ &\to\sum_{j=0}^\infty\frac{(-1)^j}{n^{2j+1}}\left(\frac{n^{2j+1}}{2j+1}+o\left(n^{2j+1}\right)\right)\\ &\to\sum_{j=0}^\infty\frac{(-1)^j}{2j+1}.\end{align}$$

  • Interesting idea. However, even $$\sum_{j=1}^\infty(-1)^jO(n^{-m})$$ cannot be simply assumed to vanish unless the $O(n^{-m})$ term is independent of $j$, and the coefficients of the Faulhaber series are not independent of $j$. I'll bet there is some way to fix this, but it may involve a bit of work. – robjohn Apr 07 '15 at 09:12
  • Yep. If I am right, the next coefficient in Faulhaber's formula is always $\frac12$, which should settle the case for this term. –  Apr 07 '15 at 10:31
  • Yes, but no other term is constant in $j$ and this causes problems. As I said, there must be a way to handle this, but it is not clear how difficult that might be. – robjohn Apr 07 '15 at 11:25
  • @robjohn: need to look at the asymptotic behavior of the Bernouilli numbers. Also the original summation should be stopped at $n-1$ to avoid the divergence at $1$. –  Apr 07 '15 at 12:32