See if series converges using the root test

[Logan Land]

((2+i)/(3-4i))^(2n)

I know it converges with 1/5 < 0

but I don't understand how to obtain 1/5
unless somehow its like this
(2+i)^2 = (4+1)=5
(3-4i)^2 = (9+19)=25
which is 5/25 which is 1/5
but that doesn't seem like the correct way to me

 

[MarkFL]

I am assuming here that:

\(\displaystyle a_n=\left(\frac{2+n}{3-4n}\right)^{2n}\)

And so:

\(\displaystyle C=\lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}=\frac{1}{16}<1\)

And so we may conclude the series converges absolutely.

 

[Logan Land]

I am assuming here that:

\(\displaystyle a_n=\left(\frac{2+n}{3-4n}\right)^{2n}\)

And so:

\(\displaystyle C=\lim_{n\to\infty}\sqrt[n]{\left|a_n\right|}=\frac{1}{16}<1\)

And so we may conclude the series converges absolutely.

why did you turn the i's into n's?
how the book says the answer is 1/5
it is a complex series

 

[MarkFL]

Oh, okay. Then I think what you did is right:

\(\displaystyle \left|\frac{2+i}{3-4i}\right|^2=\frac{2^2+1^2}{3^2+4^2}=\frac{1}{5}\)