1. Riemann sum: Limit

I do not understand how

$\displaystyle \lim_{n \rightarrow \infty} \sum_{i=1}^n \, c \frac{b-a}{n} = c(b-a) \, .$

Can somebody show the steps?

2. Re: Riemann sum: Limit

Note that $\displaystyle \sum_{i=1}^{n} \frac{1}{n} = 1$, thus we obtain

$\displaystyle \lim_{n \to \infty}\sum_{i=1}^{n} c \frac{b-a}{n} = c(b-a) \lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n} = c(b-a) \lim_{n \to \infty} 1 = c(b-a)$

3. Re: Riemann sum: Limit

Originally Posted by Siron
Note that $\displaystyle \sum_{i=1}^{n} \frac{1}{n} = 1$
Where did that come from, never seen that before. And how exactly does it equal one? The first two terms make up 1 + 0.5 = 1.5.

4. Re: Riemann sum: Limit

Originally Posted by Siron
Note that $\displaystyle \sum_{i=1}^{n} \frac{1}{n} = 1$[/tex]
That is a false statement.

$\displaystyle \lim_{n \rightarrow \infty} \sum_{i=1}^n \, c \frac{b-a}{n} = c(b-a) \, .$
That is an approximating sum for the integral $\displaystyle \int_a^b {cdx}$.
You can see that $\displaystyle \forall n,~\Delta x=\frac{b-a}{n}$ and $\displaystyle f(a+k\Delta x)\Delta x=c~\frac{b-a}{n}$