# Finding an equivalent of $u_{n}-u_{\infty}$ where $u_{n}= \sum_{k=1}^{n} \frac{n}{n^2+k^2}$

I would like to find an equivalent of

$$u_{n}-u_{\infty}=\sum_{k=1}^{n} \frac{n}{n^2+k^2}-u_{\infty}$$

Using Riemann sums, it is easy to show that:

$$u_{n} \sim \frac{\pi}{4}=u_{\infty}$$

Using integrals, we have:

$$\int_{1}^{n+1} \frac{n}{n^2+x^2} \mathrm dx \leq u_{n} \leq \int_{0}^{n} \frac{n}{n^2+x^2} \mathrm dx$$

$$\arctan(1+1/n)-\arctan(1/n) \leq u_{n} \leq \frac{\pi}{4}$$

$$\arctan(1+1/n)-\arctan(1/n)-\frac{\pi}{4} \leq u_{n} -\frac{\pi}{4}\leq 0$$

$$\arctan(1+1/n)-\arctan(1/n)= \frac{\pi}{4}+\frac{1}{2n}-\frac{1}{n}+o(1/n)=\frac{\pi}{4}-\frac{1}{2n}+o(1/n)$$

So:

$$-\frac{1}{2n}+o(1/n) \leq u_{n}-\frac{\pi}{4} \leq 0$$

However the inequality prevents from writing $$u_{n}-\frac{\pi}{4} \sim -\frac{1}{2n}$$ and numerical values seem to show that:

$$u_{n}-\frac{\pi}{4} \sim -\frac{1}{4n}$$

Where did I go wrong?

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Your numerical work indeed leads to the right conjecture $u_{n}-\frac{\pi}{4} \sim -\frac{1}{4n}$. I am feeling lazy, so to prove the result I will appeal to a standard result about $\text{TRAP}(n)$, the Trapezoidal Rule with division into $n$ equal parts. It is known that under suitable differentiability assumptions, which are amply met here, the error in $\text{TRAP}(n)$ is $O(1/n^2)$. Note that $$\text{TRAP}(n)=\sum_{k=1}^{n-1}\frac{n}{n^2+k^2} +\frac{1}{2}\left(\frac{n}{n^2}+\frac{n}{2n^2}\right).$$ Thus $$\text{TRAP}(n)=\sum_{k=1}^{n}\frac{n}{n^2+k^2} +\frac{1}{2}\left(\frac{n}{n^2}+\frac{n}{2n^2}\right)-\frac{n}{2n^2}=\sum_{k=1}^{n}\frac{n}{n^2+k^2}+\frac{1}{4n}.$$ It follows that $$\sum_{k=1}^{n}\frac{n}{n^2+k^2}=\frac{\pi}{4}-\frac{1}{4n}+O(1/n^2).$$
@Chon: You can calculate by approximately how much the term $\frac{n}{n^2+k^2}$ underestimates the area under $y=\frac{1}{1+x^2}$ from $\frac{k-1}{n}$ to $\frac{k}{n}$. If you want to do the details, it is very much like the proof of the estimate of the error in TRAP. That estimate first approximates the "local" error, that is, the difference between the area of the little trapezoid and the area under the curve from $\frac{k-1}{n}$ to $\frac{k}{n}$. The point is that $\frac{1}{1+x^2}$ is almost linear in the small interval. – André Nicolas Apr 10 '12 at 16:10