Question

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

Answer 1

@imqwerty

Answer 2

@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.

Answer 3

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 :)

Answer 4

Thanks a lot. Have a good time.

Answer 5

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.

Answer 6

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