Let's do some calculus! (34)

Calculus Level 5

Find the value of the positive constant $c_0$ such that the series given below

$\displaystyle \sum_{n=0}^{\infty} \dfrac{n!}{{(cn)}^n}$

diverges for $0 < c < c_0$ and converges for $c > c_0$ .

Clarification : $e \approx 2.71828$ denotes the Euler's number .

For more problems on calculus, click here .

e^{\frac 1e}

No such value exists for

c_0 \in \mathbb{R^+}

\frac{1}{e}

e

2 solutions

Aareyan Manzoor
Jun 12, 2017

alternatively using the ratio test $\lim_{n\to \infty} \dfrac{(n+1)!}{c^{n+1}(n+1)^{n+1}}\dfrac{c^n n^n}{n!}=\lim_{n\to \infty} \dfrac{n^n}{c(n+1)^n}=\dfrac{1}{ce}\leq 1\\ |c|\leq \dfrac{1}{e}$

First Last
May 28, 2017

Using Stirling's Approximation:

$\displaystyle\lim_{n\to\infty}\bigg|\frac{n!}{(cn)^n}\bigg|^\frac1{n}\approx\lim_{n\to\infty}\bigg|\frac{\sqrt{2\pi n}(\frac{n}{e})^n}{(cn)^n}\bigg|^\frac1{n}$

$\displaystyle\bigg|\frac1{ce}\bigg|<1\quad\text{for}\quad 0<c<\frac1{e}$

By the root test, the series converges for c greater than $\frac1{e}$ and diverges for c less than $\frac1{e}$ .

@Jasper Braun There is a typo in the last line, I think you meant ratio test.

Ankit Kumar Jain - 3 years, 1 month ago

It does not matter both the root test and the ratio test would give the same limit and hence same result. It is a consequence of Cauchy's Second limit theorem

Arghyadeep Chatterjee - 2 years, 1 month ago

Let's do some calculus! (34)

For more problems on calculus, click here .

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