Convergence of the function series ∑n!(nx)n for x&lt;0

Step 1 It says that n!∼2πn(nen as n→∞ As such, n!nn∼2πn(ne)nnn=2πnen So, you need xn→0 faster than en→∞, which is where the result comes from.

Step 1 Because x∈R, a good idea is to study the absolute convergence of the series. Let: an=|n!(x×n)n| Then, we can applay the root-criteria. In particluar, we have to determine the following limit: limn→+∞ann=limn→+∞|n!(x×n)n|n=limn→+∞n!nnn|x|=(×) Theorem: Let {bn}n≥n0 a real sequence, such that an&gt;0 at least definitely. If limn→+∞bnn and limn→+∞bn+1bn exist, then their limits are equal. Here a post where I used this tool. Let bn=n!nnn, we have: limn→+∞bn+1bn=limn→+∞(n+1)!(n+1)n+1×nnn!=limn→+∞(n+1)!(n+1)×n!×(nn+1)n=limn→+∞(1−1n)n =e−1=1e So, the first limits becomes: (×)=limn→+∞n!nnn|x|=1e×|x| By root-criteria, the series converges if and only if: 1e×|x|&lt;1⇒1e&lt;|x|⇒x∈(−∞,−1e)

Step 1 It says that n!∼2πn(nen as n→∞ As such, n!nn∼2πn(ne)nnn=2πnen So, you need xn→0 faster than en→∞, which is where the result comes from.

Step 1 Because x∈R, a good idea is to study the absolute convergence of the series. Let: an=|n!(x×n)n| Then, we can applay the root-criteria. In particluar, we have to determine the following limit: limn→+∞ann=limn→+∞|n!(x×n)n|n=limn→+∞n!nnn|x|=(×) Theorem: Let {bn}n≥n0 a real sequence, such that an&gt;0 at least definitely. If limn→+∞bnn and limn→+∞bn+1bn exist, then their limits are equal. Here a post where I used this tool. Let bn=n!nnn, we have: limn→+∞bn+1bn=limn→+∞(n+1)!(n+1)n+1×nnn!=limn→+∞(n+1)!(n+1)×n!×(nn+1)n=limn→+∞(1−1n)n =e−1=1e So, the first limits becomes: (×)=limn→+∞n!nnn|x|=1e×|x| By root-criteria, the series converges if and only if: 1e×|x|&lt;1⇒1e&lt;|x|⇒x∈(−∞,−1e)

Expert Answer to "Convergence of the function series ∑n!(nx)n for x<0"

percibaa8

Answered question

2022-01-15

Convergence of the function series

\sum \frac{n!}{{(n x)}^{n}}

for

x < 0

Answer & Explanation

Melinda McCombs

Beginner2022-01-16Added 38 answers

Step 1
It says that

n! \sim \sqrt{2 π n} ({\frac{n}{e}}^{n}

n \to \infty

As such,

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

So, you need

x^{n} \to 0

faster than

e^{n} \to \infty

, which is where the result comes from.

xandir307dc

Beginner2022-01-17Added 35 answers

Step 1
Because $x \in R$ , a good idea is to study the absolute convergence of the series. Let:
$a_{n} = | \frac{n!}{(x \times n)^{n}} |$
Then, we can applay the root-criteria. In particluar, we have to determine the following limit:
$lim_{n \to + \infty} \sqrt[n]{a_{n}} = lim_{n \to + \infty} \sqrt[n]{| \frac{n!}{(x \times n)^{n}} |} = lim_{n \to + \infty} \frac{\sqrt[n]{\frac{n!}{n^{n}}}}{| x |} = (\times)$
Theorem: Let ${b_{n}}_{n \geq n_{0}}$ a real sequence, such that $a_{n} > 0$ at least definitely. If $lim_{n \to + \infty} \sqrt[n]{b_{n}}$ and $lim_{n \to + \infty} \frac{b_{n + 1}}{b_{n}}$ exist, then their limits are equal. Here a post where I used this tool.
Let $b_{n} = \sqrt[n]{\frac{n!}{n^{n}}}$ , we have:
$lim_{n \to + \infty} \frac{b_{n + 1}}{b_{n}} = lim_{n \to + \infty} \frac{(n + 1)!}{{(n + 1)}^{n + 1}} \times \frac{n^{n}}{n!} = lim_{n \to + \infty} \frac{(n + 1)!}{(n + 1) \times n!} \times {(\frac{n}{n + 1})}^{n} = lim_{n \to + \infty} {(1 - \frac{1}{n})}^{n}$
$= e^{- 1} = \frac{1}{e}$
So, the first limits becomes:
$(\times) = lim_{n \to + \infty} \frac{\sqrt[n]{\frac{n!}{n^{n}}}}{| x |} = \frac{1}{e \times | x |}$
By root-criteria, the series converges if and only if:
$\frac{1}{e \times | x |} < 1 \Rightarrow \frac{1}{e} < | x | \Rightarrow x \in (- \infty, - \frac{1}{e})$

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