1. ## Summation

I'm having trouble figuring this one out:

$\displaystyle lim_{n->infinity}\sum\limits_{i = 1}^{\color{red}n} {\frac{{f\left( {x_i } \right)}}{n}}$

when f(x)=1/x between the interval [1,2]

I keep on getting 1/infinity and that's zero, but the correct answer is appr. .693. thank you

Note: its actually the summation of f(xsubi)*change in x but (2-1)/n is 1/n so i simplified it

2. What is .693?. It's ln(2). See now?.

3. no i don't unfortunately

4. Originally Posted by jarny
no i don't unfortunately
The limit of your summation is $\displaystyle \ln{|2|}$.

Which just SO HAPPENS to be the result of the integral:

$\displaystyle \int_{1}^{2}{\frac{1}{x}\,dx}$

In other words:

$\displaystyle \displaystyle \lim_{n \to \infty} \sum_{i = 1}^{n} \frac{f(x_i)}{n} = \int_{1}^{2}{\frac{1}{x}\,dx}$

Such that $\displaystyle x_i \in \big[1,2\big]$

In general:

$\displaystyle \displaystyle \lim_{n \to \infty} \sum_{i = 1}^{n} f(x_i)\Delta_i = \int_{a}^{b}{f(x)\,dx}$ for $\displaystyle x_i \in \big[a,b\big]$.

This is definition of a definite integral. Look it up. It's called the Riemann sum.

5. I wasn't looking for the integral, that's easy. I was trying to do it via the formula i set out. I was just trying to do it via limits and i was unable to.