Find the values of x for which the series coverges.(Enter your answer using interval notation.) ∑η=0∞(x−1)η2η Find the sum of the series fot those values of x.

Given seriez is, ∑η=0∞(x−1)η2η Simplity the seriez, ∑η=0∞2−η(x−1)η=∑η=0∞(x2−12)η This is an infinite geometric seriez with firts teem b=1 and common Ratio is, q=x2−12 By Ratio test, seriez will be commergent and when |x2−12|&lt;1 ∵−1&lt;x2−12&lt;1 ∵−12&lt;x2&lt;32 ∵−1&lt;x&lt;3 valve of x. Sum fo seriez, S=b1−q=13/2−x2=−2x−3

Resolved: "Find the values of x for which the..."

folklorahhe

Answered question

2021-12-04

Find the values of x for which the series coverges.(Enter your answer using interval notation.)

\sum_{η = 0}^{\infty} \frac{{(x - 1)}^{η}}{2^{η}}

Find the sum of the series fot those values of x.

Answer & Explanation

Charles Wee

Beginner2021-12-05Added 14 answers

Given seriez is,
$\sum_{η = 0}^{\infty} \frac{{(x - 1)}^{η}}{2^{η}}$
Simplity the seriez,
$\sum_{η = 0}^{\infty} 2^{- η} {(x - 1)}^{η} = \sum_{η = 0}^{\infty} {(\frac{x}{2} - \frac{1}{2})}^{η}$
This is an infinite geometric seriez with firts teem b=1 and common Ratio is, $q = \frac{x}{2} - \frac{1}{2}$
By Ratio test, seriez will be commergent and when $| \frac{x}{2} - \frac{1}{2} | < 1$
$∵ - 1 < \frac{x}{2} - \frac{1}{2} < 1$
$∵ - \frac{1}{2} < \frac{x}{2} < \frac{3}{2}$
$∵ - 1 < x < 3$ valve of x.
Sum fo seriez,
$S = \frac{b}{1} - q = \frac{1}{3} / 2 - \frac{x}{2} = - \frac{2}{x} - 3$

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