10.5.55 Does the following series converge or diverge?

[karush]

$\tiny{10.5.55}$
$\textsf{ Does the following series converge or diverge?}$
\begin{align*}\displaystyle
S_{n}&=\sum_{n=1}^{\infty}\frac{10^n n!n!}{(2n)!} \\
&=
\end{align*}
$\textit{ratio test?}$

 

[MarkFL]

Yes, the ratio test will work well here. (Yes)

 

[karush]

$\tiny{10.5.55}$
$\textsf{ 10.5.55 Does the following series converge or diverge?}$
\begin{align*}\displaystyle
S_{n}&=\sum_{n=1}^{\infty}\frac{10^n n!n!}{(2n)!} \\
&=
\end{align*}

$\textsf{The Ratio Test Let }$
$\displaystyle\sum ar^n$ $
\textsf{ be a series with positive terms and suppose that}$
\begin{align*}\displaystyle
&\lim_{{n}\to{\infty}} \frac{a_n+1}{a_n}=p \\
\end{align*}
$\textsf{Then}\\$
$\textsf{(a) the series converges if $p<1$}\\$
$\textsf{(b) the series diverges if $p>1$ or p is infinite}\\$
$\textsf{(c) the test is inconclusive if $p=1$}\\$
$\textsf{So}\\$
\begin{align*}\displaystyle
&=\lim_{{n}\to{\infty}}\frac{10^n n!n!+1}{(2n)^n}=p=\frac{5}{2}\\
\end{align*}

$\textit{The series diverges by the Ratio Test since the limit resulting from the test is $\displaystyle\frac{5}{2}$}$

 

[MarkFL]

Yes, I got:

$$p=\lim_{n\to\infty}\left(\frac{10(n+1)^2}{(2n+2)(2n+1)}\right)=\frac{5}{2}>1$$

Divergent. $\checkmark$

 

[alyafey22]

$$(2n)! = 1 \times 2 \times \cdots (2n-1)(2n) \leq 2 \times 2 \cdots (2n) (2n) = 4^n(n!)^2$$

Hence

$$\sum_{n=1}^\infty \frac{10^n (n!)^2}{(2n)!} \geq \sum_{n=1}^\infty \left( \frac{5}{2}\right)^n$$