"Find the radius of convergence and interval of..." solved and explained

Wribreeminsl

Answered question

2021-03-11

Find the radius of convergence and interval of convergence of the series.

\sum_{n = 1}^{\infty} \frac{x^{n}}{n 5^{n}}

Answer & Explanation

Brittany Patton

Skilled2021-03-12Added 100 answers

Consider the function as,
$a_{n} = \frac{x^{n}}{n 5^{n}}$
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,
$\Rightarrow lim_{n \to \infty} | \frac{\frac{x^{n + 1}}{(n + 1) \cdot 5^{n + 1}}}{\frac{x^{n}}{n \cdot 5^{n}}} | < 1$
$\Rightarrow lim_{n \to \infty} | \frac{x^{n + 1}}{(n + 1) \cdot 5^{n + 1}} \times \frac{n \cdot 5^{n}}{x^{n}} | < 1$
$\Rightarrow lim_{n \to \infty} | \frac{x}{(n + 1) \cdot 5} \times n | < 1$
$\Rightarrow lim_{n \to \infty} | \frac{x}{\frac{1}{n} \cdot (n + 1) \cdot 5} | < 1$
Simplify the terms further
$\Rightarrow lim_{n \to \infty} | \frac{x}{(1 + \frac{1}{n}) \cdot 5} | < 1$
$\Rightarrow | \frac{x}{(1 + 0) \cdot 5} | < 1$
$\Rightarrow | x | < 5$
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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