On an evaluation of an arctan limit
- Tolaso J Kos
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On an evaluation of an arctan limit
Evaluate the limit
$$\Omega = \lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \frac{\frac{1}{n} \arctan \left ( \frac{k}{n} \right )}{1+2\sqrt{1+\frac{1}{n} \arctan \left ( \frac{k}{n} \right )}}$$
Dan Sitaru
$$\Omega = \lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \frac{\frac{1}{n} \arctan \left ( \frac{k}{n} \right )}{1+2\sqrt{1+\frac{1}{n} \arctan \left ( \frac{k}{n} \right )}}$$
Dan Sitaru
Imagination is much more important than knowledge.
Re: On an evaluation of an arctan limit
Why not prove the more general result?
Let $f:[0, 1] \rightarrow (0, +\infty)$ be a bounded integrable function. Then:
\[\lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \frac{f\left ( \frac{k}{n} \right )}{1+2\sqrt{\frac{1}{n} f\left ( \frac{k}{n} \right )+1}} = \frac{1}{3} \int_{0}^{1} f(x) \, {\rm d}x\]
Let $f:[0, 1] \rightarrow (0, +\infty)$ be a bounded integrable function. Then:
\[\lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \frac{f\left ( \frac{k}{n} \right )}{1+2\sqrt{\frac{1}{n} f\left ( \frac{k}{n} \right )+1}} = \frac{1}{3} \int_{0}^{1} f(x) \, {\rm d}x\]
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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