Convergence Test (Ratio Test) for $\sum_n \frac{(2n)!}{n^{2n}}$

Question

I am having trouble figuring out how to tackle this problem, where I have to determine if the following sum is convergent.

$$\sum_n \dfrac{(2n)!}{n^{2n}} $$

I am assuming that you would need to apply the ratio test here, but have no idea how to continue after this:

$$\dfrac{(2n+2)!}{(n+1)^{2(n+1)}} \dfrac{n^{2n}}{(2n)!} $$

How would i go about canceling some terms out?

Answer 1

Simplifying, you get$$(2n+2)(2n+1)\frac{n^{2n}}{(n+1)^{2n+2}}=\frac{(2n+2)(2n+1)}{(n+1)^2}\left(\frac n{n+1}\right)^{2n}\to\frac4{e^2}.$$

Answer 2

One more way, if you use Stirling's approximation for $2n$: $$ (2n)! \sim \bigg(\frac{2n}{e} \Bigg)^{2n} 4 \sqrt{\pi n} $$ then $n^{2n}$ term cancels out and you are left with (set $\frac{4}{e^2} = x <1$) $$ 4 \sqrt{\pi} \sum_{k=1}^{n} \bigg(\frac{4}{e^2}\bigg)^k \sqrt{k} < 4 \sqrt{\pi} \sum_{k=1}^{n}k x^k \to \frac{ 4 \sqrt{\pi} x}{(1-x)^2} $$

this convergence works since $\frac{4}{e^2} <1$. Since the original sum is upper-bounded by the convergent sum, it converges too.

Answer 3

Lets start by stating what the ratio test is. The ratio test is given as:

$$L=\lim_{n\rightarrow \infty}\left| \frac{a_{n+1}}{a_n} \right|$$

Here $a_n$ is given as the terms of the sum, i.e:

$$\sum_n a_n$$

We can thus identify from your problem that:

$$a_n=\frac{(2n)!}{n^{2n}}$$

and (the $2$´s arrive because $n\rightarrow n+1$ gives $2n\rightarrow2n+2$),

$$a_{n+1}=\frac{(2n+2)!}{(n+1)^{2n+2}}$$

Thus giving us:

$$L=\lim_{n\rightarrow \infty}\left| \frac{\frac{(2n+2)!}{(n+1)^{2n+2}}}{\frac{(2n)!}{n^{2n}}} \right|$$

Which can be rewritten as:

$$L=\lim_{n\rightarrow \infty}\left| \frac{(2n+2)!n^{2n}}{(n+1)^{2n+2}(2n)!} \right|$$

We can note that we can rewrite this as:

$$L=\lim_{n\rightarrow \infty}\left| \frac{(2n+2)!}{(n+1)^{2}(2n)!}\frac{n^{2n}}{(n+1)^{2n}} \right|$$

Here we can identify by the following:

$$\frac{(2n+2)!}{(2n)!}=(2n+1)(2n+2)$$

Now by insertion:

$$L=\lim_{n\rightarrow \infty}\left| \frac{(2n+1)(2n+2)}{(n+1)^{2}}\frac{n^{2n}}{(n+1)^{2n}} \right| = \lim_{n\rightarrow \infty}\left| \frac{(2n+1)(2n+2)}{(n+1)^{2}}\frac{n^{2n}}{(n+1)^{2n}} \right|$$

By expanding out:

$$L = \lim_{n\rightarrow \infty}\left| \frac{4n^2 + 6n + 2}{n^2 + 2n +1}\frac{n^{2n}}{(n+1)^{2n}} \right|$$

We can see that the first fraction is just the limit of two polynomials, thus giving:

$$L = \lim_{n\rightarrow \infty}\left| 4\frac{n^{2n}}{(n+1)^{2n}} \right|$$

Now identifying the following limit:

$$e^x = \lim_{k\rightarrow \infty}\left( 1 + \frac{x}{k} \right)^k \rightarrow (e^{x})^{-1} = \lim_{k\rightarrow \infty} \left(\frac{k}{k+x} \right)^k \rightarrow e^{-1} = \lim_{k\rightarrow \infty} \left(\frac{k}{k+1} \right)^k $$

Now by insertion with $k\rightarrow n$:

$$L=\lim_{n\rightarrow \infty}\left| 4\left(\frac{1}{e^2}\right) \right| = \frac{4}{e^2}$$

Answer 4

$$\frac{(2n+2)!}{(n+1)^{2(n+1)}}\frac{n^{2n}}{(2n)!}=\frac{(2n)!(2n+1)(2n+2)}{(n+1)^{2n}(n+1)^2}\frac{n^{2n}}{(2n)!}=\frac{2(2n+1)}{n+1}(\frac{n}{n+1})^{2n}$$ $$\lim_{n\to+\infty}\frac{2(2n+1)}{n+1}=4$$ $$\lim_{n\to+\infty}(\frac{n}{n+1})^{n}=\lim_{n\to+\infty}(\frac{n+1-1}{n+1})^{n}=\lim_{n\to+\infty}((1+\frac{-1}{n+1})^{-(n+1)})^{-\frac{n}{n+1}}=\lim_{n\to+\infty}e^{-\frac{n}{n+1}}=\frac{1}{e}$$ therefore $$\frac{2(2n+1)}{n+1}(\frac{n}{n+1})^{2n} = \frac{4}{e^2} < 1$$ So the sum converges.

Answer 5

The root test is simpler.

Since $(2n)! \approx \sqrt{cn}(2n/e)^{2n}$, and $n^{1/n} \to 1$, the n-th root goes to $\dfrac{4n^2/e^2}{n^2} = 4/e^2 < 1$ so the sum converges.

If the terms are $(kn)!/n^{kn} \approx \sqrt{ckn}(kn/e)^{kn}/n^{kn} =\sqrt{ckn}(k/e)^{kn}$, the n-th root goes to $(k/e)^k$ so the sum converges for $k<e$ and diverges for $k>e$,