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Loser66
 one year ago
Find the radius of convergence of
1)\(\sum_{n=1}^\infty \dfrac{(3i)^n}{n^3}z^n\)
2) \(\sum_{n=0}^{\infty}(\dfrac{2in+1}{3n2i})^n z^n\)
Please, help
Loser66
 one year ago
Find the radius of convergence of 1)\(\sum_{n=1}^\infty \dfrac{(3i)^n}{n^3}z^n\) 2) \(\sum_{n=0}^{\infty}(\dfrac{2in+1}{3n2i})^n z^n\) Please, help

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@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.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.21st 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 xa 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 :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thanks a lot. Have a good time.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01) \[\frac{ 1 }{ R } = \lim \left \frac{a_{n+1}}{a_{n}} \right\] \[\frac{ 1 }{ R } =\lim\left \frac{ (3i)^{n+1} }{ (n+1)^{3}} \cdot \frac{ n^{3} }{ (3i)^{n} }\right\] \[\frac{ 1 }{ R } =3\cdot \lim \left (\frac{ n }{ n+1 })^{3} \right \implies R = \frac{ 1 }{ 3 }\] That one is pretty straightforward. I'll see what I can do with the 2nd one.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2np :) ws it correct? or we do something else with complex cases?
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