Mrutyunjaya M.

asked • 08/30/15# complex analysis

The radius of convergence the series

_{∝}

∑ z

^{2n }/ z^{n }is^{n=1}

a) 1 b) √2

c) √3 c) √5

ans is √2 ,but i cant understand how does it happen, please help me please, i m totally confused, please explain me in details, please sir please

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## 2 Answers By Expert Tutors

I believe the series you intended to analyze is ∑[n=1,∞] z

^{2n}/2^{n}. This can written as the power series ∑ a_{m}z^{m}where a_{m}= 2^{-}^{m/2}if m is even and 0 otherwise. Then |a_{m}|^{1/m}= 2^{-}^{1/2}if m is even and 0 otherwise. Hencelim sup |a

_{m}|^{1/m }= 2^{-1/2},and so the radius of convergence is

1/lim sup |a

_{m}|^{1/m}= 2^{1/2},which is answer b).

Roman C. answered • 08/30/15

Tutor

4.9
(636)
Masters of Education Graduate with Mathematics Expertise

The sum simplifies to ∑ z

^{n}for n=1,...,∞.Use the ratio test

a

_{n}= z^{n}so |a_{n+1}/a_{n}| = |z^{n+1}/z^{n}| = |z|Recall that if lim

_{n→∞}|a_{n+1}/a_{n}| < 1, the series converges. If it's >1 the series diverges. Inconclusive if it's =1Thus your series diverges for |z| > 1 and converges for z < 1.

The radius of convergence is therefore 1 or choice (a).

Bonus, it can be shown that this series diverges for all complex z where |z|=1 so the convergence set is the open unit disk centered at z=0.

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Mrutyunjaya M.

08/30/15