How can I give some estimation of the convergence of this series, only using comparison and majorisation of the terms?
$$\sum_{n = 0}^{+\infty} \frac{(n!)^2}{(2n)!}$$
I tried with
$$(n!)^2 = n^2\cdot (n-1)^2\cdot (n-2)^2 \ldots \leq n^2\cdot n^2\ \cdot n^2 \ldots = (n^2)^2 = n^{2n}$$
So $$\frac{(n!)^2}{(2n)!} \leq \frac{n^{2n}}{(2n)!} \qquad \text{for} n \geq 4$$
Yet I don't find a suitable majorisation for the denominator; I can only think of
$$(2n)! = 2n\cdot (2n-1)\cdot (2n-2) \ldots \geq n\cdot (n-1)\cdot (n-2) \ldots = n!$$
Hence
$$\frac{1}{(2n)!} \leq \frac{1}{n!}$$
But then I am not able to go on.
I only can/want to use comparison in this sense, and/or at the end comparison with exponential or harmonic series.
If this could be turned, somehow, into a geometric series (via some estimation / majorisation) it would be AWESOME
I cannot use Stirling or other criteria.
Not allowed to use binomial coefficient, ratio test and so on.
Thank you!
Match pairs of terms in the numerator and denominator, writing, \begin{align} \frac{(n!)^2}{(2n)!} = \left(\frac{n}{2n}\cdot\frac{n-1}{2n-2}\cdots\frac{1}{2}\right)\cdot \left( \frac{n}{2n-1}\cdot \frac{n-1}{2n-3}\cdots\frac{1}{1}\right) \end{align} where each term in the first parentheses is equal to $\frac{1}{2}$ and each term (except the last) in the second is less than $1$. Thus, for $n \geqslant 1$ \begin{align} \frac{(n!)^2}{(2n)!} \leqslant \frac{1}{2^n}. \end{align} The sum you want is then increasing (all terms are positive) and bounded above by a convergent geometric series. Thus it must also be convergent.