Find the interval of convergence of the power series. ∑n=0∞x5nn!

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rocedwrp

Answered question

2021-03-06

Find the interval of convergence of the power series.

\sum_{n = 0}^{\infty} \frac{x^{5 n}}{n!}

Answer & Explanation

avortarF

Skilled2021-03-07Added 113 answers

Consider the given power series:
$\sum a_{n} = \sum_{n = 1}^{\infty} \frac{x^{5 n}}{n!}$
Here the objective is to find the interval of x for which the given power series is convergent.
According to the ratio test
$L = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} |$
If $L < 1$ then the series converges absolutely
If $L > 1$ then the series is divergent
Here $a_{n} = \frac{x^{5 n}}{n!}$ Replace $n \to n + 1$
$a_{n + 1} = \frac{x^{5 n + 5}}{(n + 1)!}$
Use ratio test for convergence
$L = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} |$
Substitute $a_{n + 1} = \frac{x^{5 n + 5}}{(n + 1)!}$ and $a_{n} = \frac{x^{5 n}}{n!}$
$L = lim_{n \to \infty} | \frac{\frac{x^{5 n + 5}}{(n + 1)!}}{\frac{x^{5 n}}{n!}} |$
$L = lim_{n \to \infty} | \frac{n!}{x^{5 n}} \frac{x^{5 n + 5}}{(n + 1)!} |$
$L = lim_{n \to \infty} | \frac{n!}{x^{5 n}} \frac{x^{5 n} \times x^{5}}{(n + 1) (n)!} |$
$L = lim_{n \to \infty} | \frac{x^{5}}{(n + 1)} |$
$L = | x^{5} | lim_{n \to \infty} \frac{1}{(n + 1)}$
$L = | x^{5} | \times 0$
$L = 0 < 1$
Here the limit is less than 1, and independent of the value of x.
Hence the given power series is convergent for all $x \in (- \infty, \infty)$

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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