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Question 06 (b)

For which values of x does the series $\sum _{n=1}^{\infty }{\frac {2}{3^{n}(x1)^{n+1}}}$ converge?

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Hint

Use the ratio test to determine the intervals where this series converge

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Solution

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Apply the ratio test. Since we have $a_{n}={\frac {2}{3^{n}(x1)^{n+1}}}.$ and ${\frac {a_{n+1}}{a_{n}}}={\frac {2}{3^{n+1}(x1)^{n+2}}}\cdot {\frac {3^{n}(x1)^{n+1}}{2}}={\frac {1}{3(x1)}},$
the following three situations are given $\lim _{n\rightarrow \infty }\left{\frac {a_{n+1}}{a_{n}}}\right{\begin{cases}>1,&{\text{the series diverges}}\\=1,&{\text{inconclusive}}\\<1,&{\text{the series converges}}\\\end{cases}}$
Therefore, for ${\textstyle x1>{\frac {1}{3}}}$, we have ${\textstyle \lim _{n\rightarrow \infty }{\frac {a_{n+1}}{a_{n}}}={\frac {1}{3x1}}<1}$, so that the series converges.
Now, we determine the convergence of series when $x1={\frac {1}{3}}$.
When $x1={\frac {1}{3}}$ and $x1={\frac {1}{3}}$ , we have
$a_{n}={\frac {2}{3^{n}(x1)^{n+1}}}={\frac {2}{3^{n}\left({\frac {1}{3}}\right)^{n+1}}}=6$
and
$a_{n}={\frac {2}{3^{n}(x1)^{n+1}}}={\frac {2}{3^{n}\left({\frac {1}{3}}\right)^{n+1}}}=(1)^{n+1}6$, respectively.
Since in both case, $\lim _{n\to \infty }a_{n}\neq 0$, by the divergence test, the series doesn't converges.
To summarize, the series converges on ${\textstyle \color {blue}\{x:x1>{\frac {1}{3}}\}=\{x<{\frac {2}{3}}{\text{ or }}x>{\frac {4}{3}}\}.}$

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