is n!/n^n convergent?

asked by guest
on Nov 27, 2024 at 11:29 am



You asked:

Determine whether the following series converges or diverges: \(\frac{n!}{{n}^{n}}+\ldots\)

MathBot Answer:

The infinite series \(\displaystyle\sum_{n=1}^{\infty} \frac{n!}{n^{n}}\) converges.

We can use the ratio test to determine convergence of this series.

Given a series \(\displaystyle\sum a_n\), define \(L = \displaystyle\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|\)
 •If \(L < 1\) the series converges absolutely.
 •If \(L > 1\) the series diverges.
 •If \(L = 1\) or the limit does not exist the test is inconclusive.
$$\begin{aligned}L &= \lim_{n \to \infty}\left|\frac{\left(n + 1\right)^{- n - 1} \left(n + 1\right)!}{n^{- n} n!}\right| \\ &= \lim_{n \to \infty} n^{n} \left|{\frac{\left(n + 1\right)!}{n \left(n + 1\right)^{n} n! + \left(n + 1\right)^{n} n!}}\right| \\ &= e^{-1}\end{aligned}$$Therefore the series converges.

\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).

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