The problem states to find the interval of convergence of the power series. How did we get step 3 to equal 0? This is a solution from my book in case you are wondering. If you plug a big number for x like 1000(since n is going to infinity), how does it converge, or go to zero?
Ratio Test:
Formula:
\(\displaystyle \lim\limits_{n \to \infty}\dfrac{a_{n+1}}{a_{n}}\)
Problem:
\(\displaystyle \displaystyle \Sigma_{n=0}^{\infty}\dfrac{2^n(x-3)^n}{n!}\)
Steps:
1. \(\displaystyle \lim\limits_{n \to \infty}\bigg|\dfrac{\dfrac{2^{n+1}(x-3)^{n+1}}{(n+1)!}}{\dfrac{2^{n}(x-3)^n}{n!}}\bigg|<1\)
2. \(\displaystyle \lim\limits_{n \to \infty}\bigg|\dfrac{2^{n+1}\cdot(x-3)^{n+1}\cdot n!}{2^n \cdot (x-3)^n \cdot (n+1)!}\bigg|<1\)
3. \(\displaystyle \lim\limits_{n \to \infty}\bigg|\dfrac{2 \cdot (x-3)}{n+1}\bigg|=0\)
4. So the interval of convergence is \(\displaystyle (-\infty, \infty)\)
Ratio Test:
Formula:
\(\displaystyle \lim\limits_{n \to \infty}\dfrac{a_{n+1}}{a_{n}}\)
Problem:
\(\displaystyle \displaystyle \Sigma_{n=0}^{\infty}\dfrac{2^n(x-3)^n}{n!}\)
Steps:
1. \(\displaystyle \lim\limits_{n \to \infty}\bigg|\dfrac{\dfrac{2^{n+1}(x-3)^{n+1}}{(n+1)!}}{\dfrac{2^{n}(x-3)^n}{n!}}\bigg|<1\)
2. \(\displaystyle \lim\limits_{n \to \infty}\bigg|\dfrac{2^{n+1}\cdot(x-3)^{n+1}\cdot n!}{2^n \cdot (x-3)^n \cdot (n+1)!}\bigg|<1\)
3. \(\displaystyle \lim\limits_{n \to \infty}\bigg|\dfrac{2 \cdot (x-3)}{n+1}\bigg|=0\)
4. So the interval of convergence is \(\displaystyle (-\infty, \infty)\)
