Find out the answer to "For which positive integers k is the following..."

Coroware

Answered question

2021-11-21

For which positive integers k is the following series convergent?

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

Answer & Explanation

Linda Tincher

Beginner2021-11-22Added 14 answers

Step 1
Let's use the Ratio Test to check for the series; convergence:
$| \frac{a_{n + 1}}{a_{n}} | = | \frac{\frac{{((n + 1)!)}^{2}}{(k (n + 1))!}}{\frac{{(n!)}^{2}}{(k n)!}} |$
$= \frac{\frac{{((n + 1)!)}^{2}}{(k (n + 1))!}}{\frac{{(n!)}^{2}}{(k n)!}} = \frac{{((n + 1)!)}^{2}}{{(n!)}^{2}} \cdot \frac{(k n)!}{(k (n + 1))!}$
$= \frac{{(n! \cdot (n + 1))}^{2}}{{(n!)}^{2}} \cdot \frac{(k n)!}{(k n + k)!}$
$= \frac{{(n!)}^{2} \cdot {(n + 1)}^{2}}{{(n!)}^{2}} \cdot \frac{(k n)!}{(k n)! \cdot (k n + 1) \cdot (k n + 2) \cdot_{\dots} \cdot (k n + k)}$
$= \frac{{(n + 1)}^{2}}{(k n + 1) \cdot (k n + 2) \cdot_{\dots} \cdot (k n + k)}$
Now, let's examine, different values of k to find $lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} |$ . First, for $k = 1$ , we have:
$| \frac{a_{n + 1}}{a_{n}} | = \frac{{(n + 1)}^{2}}{n + 1} = n + 1 \to \infty$ ,
so the series diverges.
Now let $k = 2$
$∣ \frac{a_{n + 1}}{a_{n}} = \frac{{(n + 1)}^{2}}{(2 n + 1) (2 n + 2)} = \frac{n^{2} + 2 n + 1}{4 n^{2} + 6 n + 2} \to \frac{1}{4} < 1$
For $k > 2$ , the degree of a polynomial we get in the denominator will be k. The d

Steacensen69

Beginner2021-11-23Added 15 answers

Step 1
First we need

a_{n} \to 0

. We have

a_{n} = \frac{{(n!)}^{2}}{(k n)!} \sim \frac{2 π n {(\frac{n}{e})}^{2 n}}{\sqrt{2 π k n} {(\frac{k n}{e})}^{k n}}

= \frac{\sqrt{2 π n}}{\sqrt{k}} \frac{1}{k^{k n}} {(\frac{n}{e})}^{(2 - k) n}

Hense,

k \geq 2

the series converges by comparison test.

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