# Riemann sum: Limit

• March 24th 2013, 10:27 AM
Riemann sum: Limit
I do not understand how

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

Can somebody show the steps?
• March 24th 2013, 11:19 AM
Siron
Re: Riemann sum: Limit
Note that $\sum_{i=1}^{n} \frac{1}{n} = 1$, thus we obtain

$\lim_{n \to \infty}\sum_{i=1}^{n} c \frac{b-a}{n} = c(b-a) \lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n} = c(b-a) \lim_{n \to \infty} 1 = c(b-a)$
• March 26th 2013, 07:16 AM
Re: Riemann sum: Limit
Quote:

Originally Posted by Siron
Note that $\sum_{i=1}^{n} \frac{1}{n} = 1$

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.
• March 26th 2013, 07:31 AM
Plato
Re: Riemann sum: Limit
Quote:

Originally Posted by Siron
Note that $\sum_{i=1}^{n} \frac{1}{n} = 1$[/tex]

That is a false statement.

Quote:

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