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Riemann’s Integrals Question

I have the following question, $ \lim_{n\to\infty} \sum_{i=1}^n \tan((\frac \pi {3n})i) \times \frac \pi {3n} $

and was wondering which limit laws I could use to work out the answer? this question is derived from a riemann integral question. I know I can take out the $\frac \pi {3n} $ outside the summation so I'm left with

$ \lim_{n\to\infty} \frac \pi {3n} \sum_{i=1}^n \tan((\frac \pi {3n})i) $

but I'm not sure what to do next?

2 Answers 2

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If $f:[a,b]\to \mathbb{R}$ be continous, then

$\displaystyle \lim_{n\to\infty} \frac{b-a}{n}\sum_{i=1}^n f\left(a+\frac{b-a}{n}i\right) = \int_a ^b f(x)dx$

We can take $f(x)=\tan x$, $a=0-$, $b=\pi/3$. So this sum is equal to $\displaystyle \int_0^{\pi/3} \tan x dx.$

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    :) $ \int_0^{\pi/3} \tan x dx.$ was my original question and it asked to solve this using Riemanns Integral, hence i came to the following summation2012-12-24
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As you have asked this question you already know the answer. Here is how you come up with $f(x)=\tan x$

Let \begin{equation}\mathcal{P}=\left\{ 0=x_0$[0,1]$ We want $U_{f,\mathcal{P}}= \sum_{i=1}^n \tan((\frac \pi {3n})i)\frac \pi {3n} \iff \sum\limits_{i=1}^{n}\frac{\sup_{x\in [x_{{i-1}},x_i]}f(x)}n= \sum_{i=1}^n \tan((\frac \pi {3n})i)\frac \pi {3n} $ If we choose an increasing function this simplifies to $\sum_{i=1}^n \tan((\frac \pi {3n})i)\frac \pi {3n} =\sum_{i=1}^{n}\frac{f(x_i)}n$ Matching the terms gives $f(x_i)= \tan((\frac \pi {3n})i)\frac \pi {3}\Rightarrow f(x)= \tan((\frac \pi {3n})nx)\frac \pi {3}=\frac{\pi}3\tan(\frac{\pi x}3)$

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    i was simply wondering would it be possible to remove the $tan(\frac \pi {3n})$ outside the summation, and hence only leaving $\sum_{i=1}^n i$2012-12-24