Thread: Need Help Finding the Integral of Convergence?

1. Need Help Finding the Integral of Convergence?

hey!

first off i'm new to this forum

got a few questions on my homework that I am stuck on. I represent the summation sign with a capital E. n! means n factorial

1) E (n! x^2n / (3n-1)!)

2) E (-3/sqrt(n))(x/2)^n

thanks a bunch if you help me out!

2. Originally Posted by JhonnyBoi
hey!

first off i'm new to this forum

got a few questions on my homework that I am stuck on. I represent the summation sign with a capital E. n! means n factorial

1) E (n! x^2n / (3n-1)!)

2) E (-3/sqrt(n))(x/2)^n

thanks a bunch if you help me out!
For the first use the ratio test

$\lim_{ n \to \infty}\left| \frac{a_{n+1}}{a_n} \right|= \lim _{n\to \infty}\left| \frac{x^{2(n+1)}}{[3(n+1)-1]!}\cdot \frac{(3n-1)!}{x^{2n}}\right|=$

$\lim_{n \to \infty} \left| \frac{x^{2}}{(3n+2)(3n+1)(3n)} \right|=0$

By the ratio test the radius of convergence is all real numbers....

Try this test on the 2nd and see what happens

3. Originally Posted by JhonnyBoi
hey!

first off i'm new to this forum

got a few questions on my homework that I am stuck on. I represent the summation sign with a capital E. n! means n factorial

1) E (n! x^2n / (3n-1)!)

2) E (-3/sqrt(n))(x/2)^n

thanks a bunch if you help me out!
You mean "interval" of convergence. A fairly standard method of determining radius of convergence is to use the ratio test:
A series $\sum a_n$ converges if the ratio $|a_{n+1}/a_n|< 0$.

For 1, $a_n= \frac{n!x^{2n}}{(3n-1)!)}$ and $a_{n+1}= \frac{(n+1)! x^{2(n+1)}{(3(n+1)-1)!}= \frac{(n+1)!x^{2n}x^2}{(3n+2)!}$. The ratio is $|\frac{(n+1)!x^{2n}x^2}{(3n+2)!}\frac{(3n-1)!}{n!x^{2n}}|$ $= \frac{(n+1)!}{n!}\frac{x^{2n}}{x^{2n}\frac{(3n-1)!}{(3n+2)!}x^2$ $= \frac{n+1}{(3n+2)(3n+1)(3n)}x^2$. Take the limit as n goes to infinity, set that < 1, and solve for x.

For 2, $a_n= \frac{-3}{\sqrt{n}}(x/2)^n$, $a_{n+1}= \frac{-3}{\sqrt{n+1}}(x/2)^{n+1}$. The ratio is $|\frac{-3}{\sqrt{n+1}}(x/2)^{n+1}\frac{\sqrt{n+1}}{-3(x/2)^n}|$ $=\frac{\sqrt{n+1}}{\sqrt{n}}(x/2)$. Again, take the limit as n goes to infinity, set it < 1, and solve for x.

4. hey

first off thanks for the help and your right its interval not integral. that was my bust

Originally Posted by HallsofIvy
For 2, $a_n= \frac{-3}{\sqrt{n}}(x/2)^n$, $a_{n+1}= \frac{-3}{\sqrt{n+1}}(x/2)^{n+1}$. The ratio is $|\frac{-3}{\sqrt{n+1}}(x/2)^{n+1}\frac{\sqrt{n+1}}{-3(x/2)^n}|$ $=\frac{\sqrt{n+1}}{\sqrt{n}}(x/2)$. Again, take the limit as n goes to infinity, set it < 1, and solve for x.

when you multiply it by the reciprical wouldn't it be (-3/sqrt(n+1))(x/2)^n+1 x (sqrt(n)/-3(x/2)^n) and not what you have

thanks

5. bump