Find the interval of convergence of the power series. ∑n=0∞(3x)n(2n)!

Interval of convergence using ratio test: If limn→∞|an+1an|=L If L&lt;1 then the series is absolutely convergent If L&gt;1 then the series is absolutely divergent Given that an=(3x)n(2n)! So an+1=(3x)n+1(2n+2)! Now evaluate |an+1an|, so |an+1an|=|(3x)n+1(2n+2)!(3x)n(2n)!| =|(3x)n+1(2n+2)!(2n)!(3x)n| =|(3x)(2n+2)(2n+1)| Now L=limn→∞|an+1an| =limn→∞|(3x)(2n+2)(2n+1)| =|3x|limn→∞|1n2(2+2n)(2+1n)| L=0, Here L=0&lt;1 so series is absolutely convergent for all x. Hence interval of convergence is (−∞,∞).

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bobbie71G

Answered question

2021-02-03

Find the interval of convergence of the power series.

\sum_{n = 0}^{\infty} \frac{(3 x)^{n}}{(2 n)!}

Answer & Explanation

Laith Petty

Skilled2021-02-04Added 103 answers

Interval of convergence using ratio test: If
$lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = L$
If $L < 1$ then the series is absolutely convergent
If $L > 1$ then the series is absolutely divergent
Given that
$a_{n} = \frac{(3 x)^{n}}{(2 n)!}$
So $a_{n + 1} = \frac{(3 x)^{n + 1}}{(2 n + 2)!}$
Now evaluate $| \frac{a_{n + 1}}{a_{n}} |$ , so
$| \frac{a_{n + 1}}{a_{n}} | = | \frac{\frac{(3 x)^{n + 1}}{(2 n + 2)!}}{\frac{(3 x)^{n}}{(2 n)!}} |$
$= | \frac{(3 x)^{n + 1}}{(2 n + 2)!} \frac{(2 n)!}{(3 x)^{n}} |$
$= | \frac{(3 x)}{(2 n + 2) (2 n + 1)} |$
Now
$L = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} |$
$= lim_{n \to \infty} | \frac{(3 x)}{(2 n + 2) (2 n + 1)} |$
$= | 3 x | lim_{n \to \infty} | \frac{1}{n^{2} (2 + \frac{2}{n}) (2 + \frac{1}{n})} |$
$L = 0,$
Here $L = 0 < 1$ so series is absolutely convergent for all x.
Hence interval of convergence is $(- \infty, \infty)$ .

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