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Check whether the given series $$\sum_{n=1}^{\infty} \frac{(2^n n!)^2}{(2n+1)!}$$ converges or diverges.

I did this question by using Stirling's approximation. By using the approximation, I found out that $a_n$ behaves as $\frac{e\sqrt{\pi}}{2\sqrt{n}}$ for large $n$, where $a_n=\frac{(2^n n!)^2}{(2n+1)!}$ This means that the series diverges, but I will justify it using LCT too.

Now by setting $b_n$ as $\frac{1}{n}$, we can apply LCT to $a_n$ and $b_n$ and easily see that the limit will tend to $\infty$ as we already know the behavior of $a_n$ for large $n$. This means that $\sum a_n$ diverges.

But the thing is that I don't want to use Stirling's approximation. Is there any other method by which this question can be done? Actually, I studied Stirling's approximation on my own but my proff hasn't taught this yet.

Any help is greatly appreciated.

Mittens
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Blue Cat Blues
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2 Answers2

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Recall that $(2n)!!=2^nn!$ and $(2n-1)!!=\frac{(2n)!}{2^nn!}$; you can rewrite $a_n$ in the following way: \begin{align}a_n&=\frac{2^nn!}{(2n+1)!}\cdot2^nn! =\frac{(2n)!!}{(2n+1)!}\frac{(2n)!}{(2n-1)!!}=\frac{(2n)!!}{(2n+1)(2n)!}\frac{(2n)!}{(2n-1)!!}\\&=\frac{(2n)!!}{(2n-1)!!}\frac{1}{2n+1}.\end{align} Can you conclude?

Sine of the Time
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Let $a_n=\frac{(2^nn!)^2}{(2n+1)!}$. Then \begin{align} \left(\frac{a_{n+1}}{a_n}\right)^n&=\left(\frac{2^{2(n+1)}(n+1)!(n+1)!}{(2n+3)!}\frac{(2n+1)!}{2^{2n}n!n!}\right)^n=\Big(\frac{4(n+1)^2}{(2n+3)(2n+2)}\Big)^n\\ &=\left(\frac{4n^2+8n+4}{4n^2+10n+6}\right)^n=\left(\frac{4n^2+10n+6-2n-2}{4n^2+10n+6}\right)^n\\ &=\left(1-\frac{\tfrac{2n^2+2n}{4n^2+10n+6}}{n}\right)^n\xrightarrow{n\rightarrow\infty}e^{-1/2}>e^{-1} \end{align}

Frink's test implies that the series $\sum_n a_n$ diverges.

The same conlcussion follows by application of Raabe's test

$$n\left(1-\frac{a_{n+1}}{a_n}\right)=\frac{2n^2+2n}{4n^2+10n+6}\xrightarrow{n\rightarrow\infty}\frac12<1$$ and thus, $\sum_na_n$ diverges.

Mittens
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