find-the-region-converge-of-the-series-and-find-the-sum-n-1-n-2-n-z-1-n-

Question Number 148631 by tabata last updated on 29/Jul/21

f i n d t h e r e g i o n c o n v e r g e o f t h e s e r i e s a n d

f i n d t h e s u m \underset{n = 1}{\sum^{\infty}} \frac{n}{2^{n} {(z - 1)}^{n}} ?

Commented by tabata last updated on 29/Jul/21

? ? ? ?

Answered by mathmax by abdo last updated on 30/Jul/21

u_{n} = \frac{n}{2^{n} {(z - 1)}^{n}} \Rightarrow∣ \frac{u_{n + 1}}{u_{n}} ∣=∣ \frac{n + 1}{2^{n + 1} {(z - 1)}^{n + 1}} \times \frac{2^{n} {(z - 1)}^{n}}{n} ∣

=∣ \frac{n + 1}{2 n (z - 1)} ∣\to \frac{1}{2 ∣ z - 1 ∣} < 1 \Rightarrow 2 ∣ z - 1 ∣> 1 \Rightarrow∣ z - 1 ∣> \frac{1}{2} \Rightarrow

donc la serie converge dans Λ = {z \in C / ∣ z - 1 ∣> \frac{1}{2}}

Commented by Sozan last updated on 30/Jul/21

s i r a n d t h e s u m h o w c a n f i n d ?

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