Question

Why does the sum of the series (k!)^2/(2k)! from [1,infinity] converge?

Answer 1

\[
\frac {(k!)^2} { (2 k)!}= \frac {(k!) } { (k+1)(k+2)(k+3)\cdots(k+k)}=\\
\frac 1{k+1}\frac 2{k+2}\frac 3{k+3}\cdots \frac k{k+k}\le\\
\frac 1 2\frac 1 2\frac 1 2\cdots \frac 1 2=\frac 1 {2^k}

\]
It converges by the comparison test.