find the region converge of the series and find the sum Σ


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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
	$u_n =(n/(2^n (z−1)^n )) ⇒∣(u_(n+1) /u_n )∣=∣((n+1)/(2^(n+1) (z−1)^(n+1) ))×((2^n (z−1)^n )/n)∣ =∣((n+1)/(2n(z−1)))∣→(1/(2∣z−1∣))<1 ⇒2∣z−1∣>1 ⇒∣z−1∣>(1/2) ⇒ donc la serie converge dans Λ={z ∈C /∣z−1∣>(1/2)}$
	$u_{n} = \frac{n}{2^{n} {(z - 1)}^{n}} \Rightarrow∣ \frac{u_{n + 1}}{u_{n}} ∣=∣ \frac{n + 1}{2^{n + 1} {(z - 1)}^{n + 1}} \times \frac{2^{n} {(z - 1)}^{n}}{n} ∣$ $=∣ \frac{n + 1}{2 n (z - 1)} ∣\to \frac{1}{2 ∣ z - 1 ∣} < 1 \Rightarrow 2 ∣ z - 1 ∣> 1 \Rightarrow∣ z - 1 ∣> \frac{1}{2} \Rightarrow$ $donc la serie converge dans Λ = {z \in C / ∣ z - 1 ∣> \frac{1}{2}}$
	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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