Aloha :)
Zur Prüfung, der Konvergenz der Summe$$\sum\limits_{n=1}^\infty a_n\quad;\quad a_n\coloneqq\frac{(2n+1)!}{(3n)!}$$sollst du das Quotienten-Kriterium anwenden:
$$\phantom{=}\left|\frac{a_{n+1}}{a_n}\right|=\frac{\frac{(2(n+1)+1)!}{(3(n+1))!}}{\frac{(2n+1)!}{(3n)!}}=\frac{\frac{(2n+3)!}{(3n+3)!}}{\frac{(2n+1)!}{(3n)!}}=\frac{(2n+3)!}{(3n+3)!}\cdot\frac{(3n)!}{(2n+1)!}$$$$=\frac{\cancel{(2n+1)!}\cdot(2n+2)\cdot(2n+3)}{\cancel{(3n)!}\cdot(3n+1)\cdot(3n+2)\cdot(3n+3)}\cdot\frac{\cancel{(3n)!}}{\cancel{(2n+1)!}}=\frac{(2n+2)\cdot(2n+3)}{(3n+1)\cdot(3n+2)\cdot(3n+3)}$$$$=\frac{(2n+2)\cdot(2n+3)}{(3n+1)\cdot(3n+2)\cdot(3n+3)}=\frac{\frac{1}{n^3}\cdot(2n+2)\cdot(2n+3)}{\frac{1}{n^3}\cdot(3n+1)\cdot(3n+2)\cdot(3n+3)}$$$$=\frac{\frac{1}{n}\cdot(2+\frac2n)\cdot(2+\frac3n)}{(3+\frac1n)\cdot(3+\frac2n)\cdot(3+\frac3n)}\;\stackrel{n\to\infty}{\to}\;\;\frac{0\cdot2\cdot2}{3\cdot3\cdot3}=\frac{0}{27}=0<1\quad\checkmark$$
Das Quotienten-Kriterium ist also erfüllt, die Summe konvergiert.
