Calculate this limit by first principles

Question

I have to calculate this complex limit by first principles, ie, without using sophisticated tricks like L'Hospital's rule, Stirling formula, gamma function etc. $$ \lim_{n \to \infty} \frac{[(n+1)!]^2 (2e^{i \theta})^{2n}}{(2n)(2n+1)!} $$ where $\theta$ is a fixed real number.

Answer 1

The limit of the absolute value tends to infinity, so the expression is not convergent. Indeed, let $\displaystyle a_n={4^n[(n+1)!]^2\over 2n(2n+1)! }.$ For $n\ge 4$ we get $${a_n\over a_{n-1}}= {(n+1)^2(2n-2)\over n^2\left (2n+1\right)}=\left (1+{1\over n}\right )^2\left (1-{3\over 2n+1}\right )\\ \ge \left (1+{2\over n}\right )\left (1-{3\over 2n}\right )=1+{1\over 2n}-{3\over 2n^2}\ge 1+{1\over 2n}-{3\over 8n}=1+{1\over 8n}$$ Thus $$a_n\ge \left (1+{1\over 8n}\right ) \left (1+{1\over 8(n-1)}\right )\ldots \left (1+{1\over 8\cdot 4}\right )a_3$$ By multiplying the product and dropping nonessential positive terms we get $$a_n\ge \left [1+{1\over 8}\left ({1\over 4}+{1\over 5}+\ldots +{1\over n}\right )\right ]a_3$$ The expression inside the brackets tends to $\infty$ hence $a_n\to \infty.$

Answer 2

A couple steps to make the limit nicer:

• The $\frac{[(n+1)!]^2}{(2n+1)!}$ term looks very similar to $\frac{1}{\binom{2n+2}{n+1}}$. We can multiply by $\frac{2n}{2n+2}$ without changing the limit to get it in this form.

• The $\left(2e^{i\theta}\right)^n$ is $4^n$ with a phase term. The phase term won't converge, so it's a good guess that the limit converges to zero.

• Finally, we can multiply by four and shift $n+1\mapsto n$ to arrive at

$$\frac14\lim_{n\to\infty}\frac{4^n}{\binom{2n}{n}}.$$

This doesn't converge, however it reminds me of the Catalan/arcsin series, which look somewhat like

$$\sum\binom{2n}{n}\frac{1}{2n+1}x^n.$$

There's the more complicated $$\frac{2x^2}{1-x^2} + \frac{2x\arcsin(x)}{(1-x^2)^\frac32} = \sum_{n=1}^{\infty}\frac{(2x)^{2n}}{\binom{2n}{n}}$$

(see proofwiki) which looks so similar I believe your problem is actually supposed to be a summation rather than an integral:

$$\sum_{n=0}^{\infty}\frac{[(n+1)!]^2(2e^{i\theta})^{2n}}{2n(2n+1)!}=\frac14\sum_{n=1}^{\infty}\frac{(2e^{i\theta})^{2n}}{\binom{2n}{n}}.$$

You could crunch through from here, but $\arcsin e^{i\theta}$ looks so nice while being so ugly, so there is probably a better solution.

Answer 3

HINT: Suppose you were given the corresponding series and asked to determine whether it converges. What test do you immediately apply?