## Loser66 one year ago Find the radius of convergence of 1)$$\sum_{n=1}^\infty \dfrac{(-3i)^n}{n^3}z^n$$ 2) $$\sum_{n=0}^{\infty}(\dfrac{2in+1}{3n-2i})^n z^n$$ Please, help

1. Loser66

@imqwerty

2. Loser66

@SithsAndGiggles @oldrin.bataku I know you guys are not online, I tag and if you guys online, please give me a hand. Thanks in advance.

3. imqwerty

1st we find the L $L=\lim_{n \rightarrow \infty}\left|\frac{ a_{n+1} }{ a_{n} }\right|$ when a_{n} is nth term so to get a_{n+1} we just put n+1 in the equation on simplification we get-$L=\lim_{n \rightarrow \infty}\left| \frac{ (-3i)(z)(n)^2 }{ (n+1)^2 } \right|$$L=\left| (-3i)(z) \right|\lim_{n \rightarrow \infty}\left( \frac{ n^2 }{ (n+1)^2 } \right)$$L=\left| -3iz \right|$L<1 LHS should be of the form |x-a| and the RHS will be our radius of convergence $|-3zi|<1$dividing be |-3i|$|z \pm 0|<\frac{ 1 }{ \left| -3i \right| }$ so radius of convergence is $\frac{ 1 }{ |-3i|}$well m not so sure cause i don't knw about such complex case ..i gotta hurry i have a class so i'll see this when i come back :)

4. Loser66

Thanks a lot. Have a good time.

5. anonymous

1) $\frac{ 1 }{ R } = \lim \left| \frac{a_{n+1}}{a_{n}} \right|$ $\frac{ 1 }{ R } =\lim\left| \frac{ (-3i)^{n+1} }{ (n+1)^{3}} \cdot \frac{ n^{3} }{ (-3i)^{n} }\right|$ $\frac{ 1 }{ R } =3\cdot \lim \left| (\frac{ n }{ n+1 })^{3} \right| \implies R = \frac{ 1 }{ 3 }$ That one is pretty straightforward. I'll see what I can do with the 2nd one.

6. imqwerty

np :) ws it correct? or we do something else with complex cases?