Find the value of x for which the series converges \sum_{n=1}^\infty(x+2)^n Find the sum of the series for those values of x.

tinfoQ

tinfoQ

Answered question

2021-06-02

Find the value of x for which the series converges
n=1(x+2)n Find the sum of the series for those values of x.

Answer & Explanation

d2saint0

d2saint0

Skilled2021-06-03Added 89 answers

Consider the series n=1(x+2)n
Let an=(x+2)n, an+1=(x+2)n+1
an+1an=(x+2)n+1(x+2)n
=x+2
limnan+1an=limn(|x+2|)=|x+2|
Using ratio test, this will converge, when
|x+2|<1
1<(x+2)<1
3
The series converges at values that fall within the interval (-3,-1)
And n=1(x+2)n=(x+2)1+(x+2)2+(x+2)3+...+(x+2)
=x+21(x+2)
=x+21x

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