FrozenDragon427
New member
- Joined
- Sep 17, 2020
- Messages
- 14
CThere is no reason to complicate matters.
If \(0<r<1\) then \(0<r^{n+1}<r^n<1\) i.e. If \(0<r<1\) then \(r^n\) forms a decreasing sequence and has limit \(0\).
\(\begin{gathered}
{S_N} = \sum\limits_{n = 0}^N {a{r^n}} = a + ar + a{r^2} + \cdots + a{r^N} \hfill \\
r{S_N} = ar + a{r^2} + \cdots + a{r^N} + a{r^{N + 1}} \hfill \\
{S_N} - r{S_N} = a + ar + a{r^2} + \cdots + a{r^{N + 1}} \hfill \\
{S_N} = \frac{{a - {r^{N + 1}}}}{{1 - r}}\quad \mathop {\lim }\limits_{N \to \infty } {S_N} = \frac{a}{{1 - r}} \hfill \\ \end{gathered} \)
So in #40 what is the answer?
Sure. Can you please look at my post #3 and conclude what lim(n=>infinity) (x^n) equals if 0<x<1? Just think what happens to a number between 0 and 1 when you keep multiplying it by itself. Just take a calculator, type in .5 and keep hitting the times button. Try it with .714, .001, etc. What do you eventually get?Can someone help me?