Question

Find the radius of convergence and interval of convergence of the series. sum_{n=1}^inftyfrac{x^n}{n5^n}

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ANSWERED
asked 2021-03-11
Find the radius of convergence and interval of convergence of the series.
\(\sum_{n=1}^\infty\frac{x^n}{n5^n}\)

Answers (1)

2021-03-12

Consider the function as,
\(a_n=\frac{x^n}{n5^n}\)
So it implies that,
\(a_{n+1}=\frac{x^{n+1}}{(n+1)5^{n+1}}\)
Apply the ratio test, for a converging series,
\(\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|<1\)
Substitute the values and simplify,
\(\Rightarrow\lim_{n\to\infty}|\frac{\frac{x^{n+1}}{(n+1)\cdot5^{n+1}}}{\frac{x^n}{n\cdot5^n}}|<1\)
\(\Rightarrow\lim_{n\to\infty}|\frac{x^{n+1}}{(n+1)\cdot5^{n+1}}\times\frac{n\cdot5^n}{x^n}|<1\)
\(\Rightarrow\lim_{n\to\infty}|\frac{x}{(n+1)\cdot5}\times n|<1\)
\(\Rightarrow\lim_{n\to\infty}|\frac{x}{\frac{1}{n}\cdot(n+1)\cdot5}|<1\)
Simplify the terms further
\(\Rightarrow\lim_{n\to\infty}|\frac{x}{(1+\frac{1}{n})\cdot5}|<1\)
\(\Rightarrow|\frac{x}{(1+0)\cdot5}|<1\)
\(\Rightarrow|x|<5\)
Thus, the radius of convergence of the given series is 5 units.
Consider the inequality as,
\(|x|<5\)
\(\Rightarrow-5\)
or \(x\in(-5,5)\)
Thus, the interval of convergence is (−5, 5).

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