Radius of Convergence
- Thread starter |m|
- Start date
Hello, |m|!
Did you try the Ratio Test?
. . \(\displaystyle \L\sum^{\infty}_{n=1}\frac{n^2}{5^n}\,x^n\)
\(\displaystyle \L\frac{a_{n+1}}{a_n} \:=\:\frac{(n+1)^2x^{n+1}}{5^{n+1}}\,\cdot\,\frac{5^n}{n^2\cdot x^n} \;=\;\frac{(n+1)^2}{n^2}\cdot\frac{x}{5}\)
Divide top and bottom by \(\displaystyle n^2:\) \(\displaystyle \;\L\frac{\frac{(n+1)^2}{n^2}}{\frac{n^2}{n^2}}\cdot\frac{x}{5} \:=\:\frac{\left(\frac{n+1}{n}\right)^2}{1}\cdot\frac{x}{5}\:=\:\left(1\,+\,\frac{1}{n}\right)^2\cdot\frac{x}{5}\)
Take the limit: \(\displaystyle \L\:\lim_{n\to\infty}\left|\left(1\,+\,\frac{1}{n}\right)\cdot\frac{x}{5}\right| \:=\:\left|(1\,+\,0)\cdot\frac{x}{5}\right| \:=\:\left|\frac{x}{5}\right|\)
For convergence: \(\displaystyle \L\;\left|\frac{x}{5}\right| \:<\:1\;\;\Rightarrow\;\;-1\:<\:\frac{x}{5}\:<\:1\;\;\Rightarrow\;\;-5\:<\:x\:<\:5\)
Therefore, the radius of convergence is: \(\displaystyle \L\,r\,=\,5\)
Did you try the Ratio Test?
. . \(\displaystyle \L\sum^{\infty}_{n=1}\frac{n^2}{5^n}\,x^n\)
\(\displaystyle \L\frac{a_{n+1}}{a_n} \:=\:\frac{(n+1)^2x^{n+1}}{5^{n+1}}\,\cdot\,\frac{5^n}{n^2\cdot x^n} \;=\;\frac{(n+1)^2}{n^2}\cdot\frac{x}{5}\)
Divide top and bottom by \(\displaystyle n^2:\) \(\displaystyle \;\L\frac{\frac{(n+1)^2}{n^2}}{\frac{n^2}{n^2}}\cdot\frac{x}{5} \:=\:\frac{\left(\frac{n+1}{n}\right)^2}{1}\cdot\frac{x}{5}\:=\:\left(1\,+\,\frac{1}{n}\right)^2\cdot\frac{x}{5}\)
Take the limit: \(\displaystyle \L\:\lim_{n\to\infty}\left|\left(1\,+\,\frac{1}{n}\right)\cdot\frac{x}{5}\right| \:=\:\left|(1\,+\,0)\cdot\frac{x}{5}\right| \:=\:\left|\frac{x}{5}\right|\)
For convergence: \(\displaystyle \L\;\left|\frac{x}{5}\right| \:<\:1\;\;\Rightarrow\;\;-1\:<\:\frac{x}{5}\:<\:1\;\;\Rightarrow\;\;-5\:<\:x\:<\:5\)
Therefore, the radius of convergence is: \(\displaystyle \L\,r\,=\,5\)