3

I am hoplessly trying the ratio test. So if $a_n=n!^{\frac{1}{n}}$ then we compute the limit of the ratio $\frac{a_{n+1}}{a_n}$ if it turns out to be less than 1 it converges. So $$\frac{(n+1)!^{\frac{1}{n+1}}}{n!^{\frac{1}{n}}}=(n!)^{\frac{1}{n+1}-\frac{1}{n}}(n+1)^{\frac{1}{n+1}}$$.

Now I know that $(n+1)^{\frac{1}{n+1}}$ converges to 1. How do I proceed?

Miz
  • 2,765
  • 2
    Apply Stirling's formula (https://en.wikipedia.org/?title=Stirling%27s_approximation) and you're good to go, the usual ratio test won't lead you anywhere (as far as I remember) – b00n heT Jun 14 '15 at 06:19
  • @ b00n heT So by Stirling approximation $n!$ is asymptotically similar to $\sqrt{2 \pi n}(\frac{n}{e})^n$. So I can conclude that $n!^{\frac{1}{n}}$ has the same asymptotic behavior as $\Big{(}\sqrt{2 \pi n}(\frac{n}{e})^n\Big{)}^{\frac{1}{n}}$. Of course the latter goes to infinity. Therefore so does $n!^{\frac{1}{n}}$. Is that fine? – Miz Jun 14 '15 at 06:27
  • I don't think it converges. Use the term test, i.e., show that $\lim_{n\to\infty}n!^{1/n}=\infty$ – Prasun Biswas Jun 14 '15 at 06:28
  • 1
    Yes @AloysiusGodinho. – b00n heT Jun 14 '15 at 06:29
  • Cool @ b00n heT. Thanks for the hint. But I am inquisitive Is there a standard technique that I can apply to functions like these. – Miz Jun 14 '15 at 06:31
  • Duplicate of http://math.stackexchange.com/questions/706461/calculating-the-limit-limn1-n – Barry Cipra Jun 14 '15 at 22:46

3 Answers3

6

(I added a more general result at the end.)

Here is a completely elementary proof that does not need Stirling's formula or even logs:

$n! =\prod_{k=1}^n k >\prod_{k=\lfloor n/2 \rfloor}^n k >\lfloor n/2 \rfloor^{\lceil n/2 \rceil} \ge\lfloor n/2 \rfloor^{ n/2} $.

Therefore, $n!^{1/n} >\lfloor n/2 \rfloor^{1/2} $ which diverges.

Note: You can be more precise and get better estimates, but this is enough to show divergence.

More generally, if you take the terms after $an$, where $0 < a < 1$, there are $n(1-a)$ terms, each at least $an$, so $n! >(an)^{n(1-a)} $ or $n!^{1/n} >(an)^{1-a} $.

This does not get to the true value of $cn$ for some $c$, but it gets close.

Another tack:

Divide $1$ to $n$ into $k$ parts. Each part has $n/k$ numbers with the smallest of the $j^{th}$ part being $nj/k$, for $j$ from $0$ to $k-1$.

Therefore, skipping the first part, $n! >\prod_{j=1}^{k-1} (nj/k)^{n/k} =(n/k)^{n(k-1)/k}\prod_{j=1}^{k-1} j^{n/k} =(n/k)^{n(k-1)/k}(k-1)!^{n/k} $ so $n!^{1/n} >(n/k)^{(k-1)/k}(k-1)!^{1/k} $.

For example, if $k=3$, then $n!^{1/n} >2^{1/3}(n/3)^{2/3} $.

If we take $k = \sqrt{n}$, this shows that $n!^{1/n} >(\sqrt{n})^{1-1/\sqrt{n}}(\sqrt{n}-1)!^{1/\sqrt{n}} $, or $n! >(\sqrt{n})^{\sqrt{n}-1}(\sqrt{n}-1)!^{\sqrt{n}} $.

Don't know if this is any use, but it's fun.

marty cohen
  • 110,450
3

Hint.

You can either use Stirling approximation as suggested. An alternate way if you're not supposed to know Stirling approximation is to go through logarithm:

$$\ln a_n = \frac{1}{n}\sum_{k=1}^n \ln k$$ and then using that logarithm is increasing to compare the sum to an integral.

mathcounterexamples.net
  • 71,758
  • 1
    Thanks I guess taking logarithm in situations involving products like this is the best way to gauge the growth of the sequence. – Miz Jun 14 '15 at 06:35
  • 1
    I'm guessing you are asking me to compare the sum $\sum\limits_{k=1}^n Log(k)$ with the integral of the same function from $1 \rightarrow k$ right? since they both converge or diverge together. – Miz Jun 14 '15 at 06:41
  • 1
    Yes that is right. For $n \ge 1$ integer you have $\ln x \ge \ln n$ for $x \in [n,n+1]$. – mathcounterexamples.net Jun 14 '15 at 06:48
2

Let's follow what @b00n heT has suggested:

$$n!\sim \sqrt{2\pi n}\left({n\over e}\right)^n$$

So we are left with finding the limit:

$$\lim_{n\to\infty}\left(2\pi n\right)^{1\over 2n}\left({n\over e}\right)$$

Let's look at the two terms of the product

$$\left(2\pi n\right)^{1\over 2n}\to 1$$ $${n\over e}\to \infty$$

So our sequence goes to $+\infty$

marwalix
  • 17,045