Question

Find the interval of convergence of the power series, where c > 0 and k is a positive integer. sum_{n=1}^inftyfrac{n!(x-c)^n}{1cdot3cdot5cdot...cdot(2n-1)}

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asked 2020-11-01
Find the interval of convergence of the power series, where c > 0 and k is a positive integer.
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{n}!{\left({x}-{c}\right)}^{{n}}}}{{{1}\cdot{3}\cdot{5}\cdot\ldots\cdot{\left({2}{n}-{1}\right)}}}}\)

Answers (1)

2020-11-02

To determine:
The interval of the convergence of the power series, where c > 0 and k is a positive integer.
\(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{n}!{\left({x}-{c}\right)}^{{n}}}}{{{1}\cdot{3}\cdot{5}\cdot\ldots\cdot{\left({2}{n}-{1}\right)}}}}\)
Calculation:
Let \(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{n}}{\left({x}-{c}\right)}^{{n}}={\sum_{{{n}={1}}}^{\infty}}{\frac{{{n}!{\left({x}-{c}\right)}^{{n}}}}{{{1}\cdot{3}\cdot{5}\cdot\ldots\cdot{\left({2}{n}-{1}\right)}}}}\)
By comparing
\(\displaystyle{a}_{{n}}={\frac{{{n}!}}{{{1}\cdot{3}\cdot{5}\cdot\ldots\cdot{\left({2}{n}-{1}\right)}}}}\) By using the ratio test,
\(\displaystyle{L}=\lim_{{{n}\to\infty}}{\left|{\frac{{{a}_{{{n}+{1}}}}}{{{a}_{{n}}}}}\right|}={\left|{\frac{{{\left({n}+{1}\right)}!{\left({x}-{c}\right)}^{{{n}+{1}}}}}{{{1}\cdot{3}\cdot{5}\cdot\ldots\cdot{\left({2}{n}+{1}\right)}}}}\times{\frac{{{1}\cdot{3}\cdot{5}\cdot\ldots\cdot{\left({2}{n}-{1}\right)}}}{{{n}!}}}\right|}\)
\(\displaystyle{L}=\lim_{{{n}\to\infty}}{\left|{\frac{{{\left({n}+{1}\right)}{\left({2}{n}-{1}\right)}{\left({x}-{c}\right)}}}{{{\left({2}{n}+{1}\right)}}}}\right|}\)
\(\displaystyle{L}={\left|{x}-{c}\right|}\lim_{{{n}\to\infty}}{\left|{\frac{{{\left({n}+{1}\right)}{\left({2}{n}-{1}\right)}}}{{{\left({2}{n}+{1}\right)}}}}\right|}\)
But \(\displaystyle\lim_{{{n}\to\infty}}{\left|{\frac{{{\left({n}+{1}\right)}{\left({2}{n}-{1}\right)}}}{{{2}{n}+{1}}}}\right|}=\infty\)
So, the radius of convergence is 0 and the power series converges at x = c
\(\displaystyle\Rightarrow{\left|{x}-{c}\right|}{<}{0}\)
\(\displaystyle\Rightarrow-{c}{<}{x}{<}{c}\)
But c >0
So, it is 0 Therefore, the interval of the convergence for the given power series is (0, c]

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