Question

Find the radius of convergence

\[ y= C_0\sum_{0}^{\infty}x^{2n} + C_1\sum_{0}^{\infty}x^{2n+1} \]

I keep getting stuck on this part of a problem, can someone work it out for me and show all the steps.
Thanks!

Answer 1

the constants are unimportant.
the radius of convergence of \(\sum_{n=0}^{\infty}x^n\) is \(|x|<1\) as it is a geometric series
i don't think it makes any difference that you split off the even and odd exponents, unless i am missing something

Answer 2

Well the book is saying that the answer is 1. I keep getting 0 (1/infty) when I do it.

Answer 3

can you show your work?

Answer 4

\[ \lim_{n \rightarrow \infty} | \frac{C_n}{C_{n+1}} | \]
\[ \lim_{n \rightarrow \infty} | \frac{x^{2n} + x^{2n+1}}{x^{2n+2} + x^{2n+3}} | \]
\[ \lim_{n \rightarrow \infty} | \frac{x^{2n}(1 + x)}{x^{2n}( x^2+ x^{3})} | \]
\[ \lim_{n \rightarrow \infty} | \frac{1 + x}{x^2+ x^{3}} | = 0\]

Answer 5

BTW the book shows that this should be \[ \frac{c_0 + c_1x}{1-x^2}\] Which I don't understand how they got that.

Answer 6

It's not too bad:
\[c_0 \sum_{n=0}^{\infty}x^{2n}+c_1 \sum_{n=0}^{\infty}x^{2n+1}=(c_0 + c_1 x)\sum_{n=0}^{\infty}x^{2n}\]
If |x| < 1 then:
\[\sum_{m=0}^{\infty}x^m=\frac{1}{1-x} \textrm{with }\sum_{m=0}^{\infty} \left( x^2\right)^m=\frac{1}{1-x^2} \textrm{for } |x|<1\]
Giving:
\[\frac{c_0+c_1 x}{1-x^2}\]

Answer 7

Oh wow, thank you!