Question

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.

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asked 2021-06-02
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.

Expert Answers (1)

2021-06-03

Consider the series \(\sum_{n=1}^\infty(x+2)^n\)
Let \(a_n=(x+2)^n,\ a_{n+1}=(x+2)^{n+1}\)
\(\frac{a_{n+1}}{a_n}=\frac{(x+2)^{n+1}}{(x+2)^n}\)
\(=x+2\)
\(\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}(|x+2|)=|x+2|\)
Using ratio test, this will converge, when
\(|x+2|<1\)
\(-1<(x+2)<1\)
\(-3\)
So, the series is converges at the values lies in the interval, (-3,-1)
And \(\sum_{n=1}^\infty(x+2)^n=(x+2)^1+(x+2)^2+(x+2)^3+...+(x+2)^\infty\)
\(=\frac{x+2}{1-(x+2)}\)
\(=\frac{x+2}{-1-x}\)

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