Determine if the series a n = n ! ( 1 + n ) n x n converges

Question

Determine if the series      a    n    =            n      !              (      1      +      n              )        n                  x    n  converges

Asdrubali2r · Accepted Answer

I suppose that you are asking about the radius of convergence of

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

To check for convergence, we can apply the ratio test,

\begin{aligned} | \frac{a_{n + 1}}{a_{n}} | = | \frac{\frac{(n + 1)! x^{n + 1}}{(1 + (n + 1))^{n + 1}}}{\frac{n! x^{n}}{(1 + n)^{n}}} | = | x | | \frac{(1 + n) (1 + n)^{n} n!}{(2 + n)^{n} (2 + n) n!} | \underset{n \to + \infty}{⟶} | x | e^{\frac{- n - 1}{n + 2}} \underset{n \to + \infty}{⟶} e^{- 1} | x | \end{aligned}

Hence convergence absolutely if

| x | < e

by the ratio test and divergence for

| x | > e

. If

| x | = e

so

x = - e

or

x = e

we need to test these values of x separately.
Stirling's approximation give:

n! \underset{+ \infty}{\sim} \sqrt{2 π n} (n / e)^{n}

then

\frac{n! e^{n}}{(n + 1)^{n}} \underset{+ \infty}{\sim} \sqrt{2 π n} {(\frac{n}{e})}^{n} \frac{e^{n}}{(n + 1)^{n}} = \sqrt{2 π n} \frac{n^{n}}{(n + 1)^{n}}

Hence,

\frac{\sqrt{2 π n} \frac{n^{n}}{(n + 1)^{n}}}{\sqrt{n}} = \sqrt{2 π} \frac{n^{n}}{(n + 1)^{n}} \underset{n \to + \infty}{⟶} \frac{\sqrt{2 π}}{e}

Since

0 < \frac{\sqrt{2 π}}{e} < + \infty

and the series of

\sum_{n = 1}^{+ \infty} \sqrt{n}

diverges so

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

diverges by limit comparison test.

Determine if the series a n </msub> = n ! </mrow>

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