# Convergence of $\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}x^n$

I was given the following question :

Determine the radius of convergence of the following series : $$\displaystyle\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}x^n$$

I was able to determine using the Cauchy–Hadamard formula that $$R = 4$$, but I still need to determine whether the series converges for $$x = 4$$ or $$x=-4$$.

I tried using the ratio test (for $$x = 4$$) but the result is $$1$$, so I can't really do anything with it. Any ideas on how I could approach this? Thanks!

• Ratio and root tests will be inconclusive at the endpoints (without the more advanced statements you can find on Wikipedia) because they correspond to $L=1$. Commented Jul 16 at 0:19

This is perhaps overkill but I think the maths is cute and it doesn't require any appeals to Stirling's approximation etc:

For $$x = 4$$,

$$\displaystyle\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}2^{2n}= \displaystyle\sum\limits_{n=1}^{\infty} \frac{(2n!!)^2}{(2n)!}$$

$$\displaystyle\frac{(2n!!)^2}{(2n)!} = \frac{((2n)(2n-2)(2n-4)...(2))^2}{(2n)(2n-1)(2n-2)...(2)} = \frac{(2n)(2n-2)(2n-4)...(2)}{(2n-1)(2n-3)(2n-5)...(1)} = \prod_{i=1}^n{\frac{2i}{2i-1}}$$

Now observe that $$\displaystyle\prod_{i=1}^n{\frac{2i}{2i-1}} \ge 1$$ $$\forall$$ $$n$$

And hence $$\frac{(2n!!)^2}{(2n)!} \ge 1$$ $$\forall$$ $$n$$

$$\displaystyle\sum\limits_{n=1}^{k} \frac{(n!)^2}{(2n)!}2^{2n} \ge k$$ and so certainly does not converge

For $$x = -4$$, notice that the terms in the sum are exactly the same as before but with alternating signs. Furthermore:

$$m > n \implies \displaystyle\prod_{i=1}^m{\frac{2i}{2i-1}} > \displaystyle\prod_{i=1}^n{\frac{2i}{2i-1}}$$

Hence the terms of the series are increasing in absolute value and therefore do not converge to $$0$$. Thus the series does not converge (this reasoning equally applies to $$x=4$$).

• If the terms do not tend to zero, the series cannot converge. Convergence of the terms to zero is a necessary condition for convergence of the series, and it has nothing to do with the alternating series test.
– Gary
Commented Jul 15 at 22:29
• Lol, I’ll delete that last sentence. Thanks Commented Jul 16 at 0:11
• A friendly reminder that if you feel like your question has been sufficiently answered, you can accept an answer. This awards a green tick to let other users know the answer has been accepted and awards reputation to the author which helps incentivize further contributions. Commented Jul 27 at 5:02

Stirling's approximation (good for intuition, even if it's not acceptable for an explanation) gives you $$\frac{4^n(n!)^2}{(2n)!} \approx \frac{4^n(\sqrt{2\pi n} n^n e^{-n})^2} {\sqrt{2\pi(2n)} (2n)^{2n} e^{-2n}} = \frac{4^n(2\pi n) n^{2n} e^{-2n}}{2\sqrt{\pi n} 2^{2n} n^{2n} e^{-2n}} = \sqrt{\pi n}$$ Which would tell you the terms diverge individually and the series can't converge. Wolfram Alpha seems to agree.

It looks like the terms increase, in fact, and you can show this with the ratio of consecutive terms: $$\frac{\frac{4^{n+1}((n+1)!)^2}{(2(n+1))!}}{\frac{4^n(n!)^2}{(2n)!}} = \frac{4(n+1)^2}{(2n+2)(2n+1)} = \frac{2n+2}{2n+1} > 1$$

We have $$4^n=(1+1)^{2n}>{2n\choose n}={(2n)!\over ( n!)^2}$$ Hence for $$x=\pm 4$$ the absolute value of the general term is greater than $$1,$$ which excludes the convergence.

Remark 1 The radius of convergence is equal $$4$$ and can be easily determined by the $$n$$th root test basing on the inequality $${4^n\over 2n+1}< {2n\choose n }<4^n\quad (*)$$ as the $$n$$-th term in the Newton triangle is the greatest in the $$2n$$-th row.

Remark 2 Another series $$\sum_{n=1}^\infty {(2n)!\over (n!)^2}x^n$$ is more interesting. By $$(*)$$ the radius of convergence is equal $$1/4.$$ For $$x=1/4$$ the series is divergent due to the first inequality in $$(*)$$ which gives the lower bound by $$(2n+1)^{-1}.$$ For $$x=-1/4$$ the series is convergent by the Leibniz test as the sequence $${(2n)!\over 4^n(n!)^2}$$ is decreasing and convergent to $$0.$$

• The radius of convergence is $4$, not $\tfrac14$.
– J.G.
Commented Jul 16 at 9:11
• @J.G. The radius of the series in OP is equal 4. The one for the series in Remark 2 is equal $1/4.$ The reciprocals of the coefficients of the OP series is considered there. Commented Jul 16 at 9:46
• Incidentally, Remark 2's series is $(1-4x)^{-1/2}-1$ when it converges.
– J.G.
Commented Jul 16 at 10:36
• @J.G. That's true as the coefficients are equal $4^n{-1/2\choose n}.$ Still the convergence at $x=-1/4$ requires a proof and then the value of the sum $2^{-1/2}$ can be determined on the basis of Abel's theorem on power series. Commented Jul 16 at 11:45