Find the interval of convergence of the following series. frac{(x-4)^n}{3^n}

Question
Series
asked 2021-02-22
Find the interval of convergence of the following series.
\(\displaystyle{\frac{{{\left({x}-{4}\right)}^{{n}}}}{{{3}^{{n}}}}}\)

Answers (1)

2021-02-23
Given the series:
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left({x}-{4}\right)}^{{n}}}}{{{3}^{{n}}}}}\)
The nth term is given as:
\(\displaystyle{a}_{{n}}={\frac{{{\left({x}-{4}\right)}^{{n}}}}{{{3}^{{n}}}}}\)
Applying the ratio test, we get:
\(\displaystyle{L}=\lim_{{{n}\to\infty}}{\left|{\frac{{{a}_{{{n}+{1}}}}}{{{a}_{{n}}}}}\right|}\)
\(\displaystyle\Rightarrow{L}=\lim_{{{n}\to\infty}}\le{f}{t}{\left|{\frac{{{\frac{{{\left({x}-{4}\right)}^{{{n}+{1}}}}}{{{3}^{{{n}+{1}}}}}}}}{{{\frac{{{\left({x}-{4}\right)}^{{n}}}}{{{3}^{{n}}}}}}}}{r}{i}{g}{h}{t}\right|}\)
\(\displaystyle\Rightarrow{L}=\lim_{{{n}\to\infty}}{\left|{\frac{{{\left({x}-{4}\right)}^{{{n}+{1}}}}}{{{3}^{{{n}+{1}}}}}}\times{\frac{{{3}^{{n}}}}{{{\left({x}-{4}\right)}^{{n}}}}}\right|}\)
\(\displaystyle\Rightarrow{L}=\lim_{{{n}\to\infty}}{\left|{\frac{{{\left({x}-{4}\right)}^{{{n}+{1}-{n}}}}}{{{3}^{{{n}+{1}-{n}}}}}}\right|}\)
\(\displaystyle\Rightarrow{L}=\lim_{{{n}\to\infty}}{\left|{\frac{{{\left({x}-{4}\right)}}}{{{3}}}}\right|}\)
Now plugging the limit, we get:
\(\displaystyle\Rightarrow{L}={\left|{\frac{{{\left({x}-{4}\right)}}}{{{3}}}}\right|}\)
For series to converge:
\(\displaystyle{\left|{\frac{{{\left({x}-{4}\right)}}}{{{3}}}}\right|}{<}{1}\)</span>
\(\displaystyle{\left|{x}-{4}\right|}{<}{3}\)</span>
Hence the radius of convergence is:
\(\displaystyle{R}={3}\)
The interval of convergence is:
\(\displaystyle-{3}{<}{x}-{4}{<}{3}\)</span>
\(\displaystyle-{3}+{4}{<}{x}{<}{3}+{4}\)</span>
\(\displaystyle{1}{<}{x}{<}{7}\)</span>
Now checking for x=1, we get series:
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left({x}-{4}\right)}^{{n}}}}{{{3}^{{n}}}}}={\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{n}}}}{{{3}^{{n}}}}}\)
The above series converges since \(\displaystyle\sum{\left|{a}_{{n}}\right|}=\sum{\frac{{{1}}}{{{3}^{{n}}}}}\) converges by geometric series test.
Now checking for \(\displaystyle{x}={7}\), we get series:
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left({x}-{4}\right)}^{{n}}}}{{{3}^{{n}}}}}={\sum_{{{n}={1}}}^{\infty}}{\frac{{{\left({3}\right)}^{{n}}}}{{{3}^{{n}}}}}={\sum_{{{n}={1}}}^{\infty}}{1}\)
The above is a diverging series.
Hence the interval of convergence is: \(\displaystyle{1}\leq{x}{<}{7}\)</span>
Final Answer:
Interval of convergence of the series is :
\(\displaystyle{1}\leq{x}{<}{7}\)</span>
0

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