Solved: "Determine whether the following series converges. ∑k=1∞9(4k)!(k!)4"

Sinead Mcgee

Answered question

2021-02-14

Determine whether the following series converges.

\sum_{k = 1}^{\infty} \frac{9 (4 k)!}{(k!)^{4}}

Answer & Explanation

okomgcae

Skilled2021-02-15Added 93 answers

Consider the given series:
$\sum_{k = 1}^{\infty} \frac{9 (4 k)!}{(k!)^{4}}$
Now, let
$a_{k} = \frac{9 (4 k)!}{(k!)^{4}}$
Now, apply ratio test:
First evaluate the following limit:
$r = lim_{k \to \infty} \frac{a_{k + 1}}{a_{k}} = lim_{k \to \infty} \frac{\frac{9 (4 (k + 1))!}{[(k + 1)!]^{4}}}{\frac{9 (4 k)!}{(k!)^{4}}}$
$= lim_{k \to \infty} \frac{(4 k + 4)!}{(4 k)!} \cdot \frac{(k!)^{4}}{[(k + 1)!]^{4}}$
$= lim_{k \to \infty} \frac{(4 k + 4) (4 k + 3) (4 k + 2) (4 k + 1) (4 k)!}{(4 k)!} \cdot [\frac{(k!)}{(k + 1) (k!)}]^{4}$
$= lim_{k \to \infty} \frac{(4 k + 4) (4 k + 3) (4 k + 2) (4 k + 1)}{(k + 1)^{4}}$
$= lim_{k \to \infty} \frac{(4 k + \frac{4}{k}) (4 k + \frac{3}{k}) (4 k + \frac{2}{k}) (4 k + \frac{1}{k})}{(1 + \frac{1}{k})^{4}}$
$= (4) (4) (4) (4)$
$= 256$
Since, $256 > 1$
Therefore, by ratio test series is divergent.
Hence, the ratio test yields $r = 256$ . This is greater than 1, so the series diverges by the ratio test.

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