# limits and summation [duplicate]

Possible Duplicate:
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?

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## marked as duplicate by Nameless, Davide Giraudo, Hagen von Eitzen, tomasz, Matthew PresslandDec 24 '12 at 15:34

This question was marked as an exact duplicate of an existing question.

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 summation – jill Dec 24 '12 at 11:46

As you have asked this question you already know the answer. Here is how you come up with $f(x)=\tan x$

Let $$\mathcal{P}=\left\{ 0=x_0<x_1<...<\frac{i}{n}<...<x_n=1 \right\}$$ partition $[0,1]$ 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 $\sum_{i=1}^n i$ – jill Dec 24 '12 at 11:55