Aloha :)
$$\pink{\text{zu a) }}\quad \sum\limits_{n=1}^\infty a_n\cdot x^n\quad;\quad a_n\coloneqq\frac{1}{n^2}$$$$r=\lim\limits_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|=\lim\limits_{n\to\infty}\frac{\frac{1}{n^2}}{\frac{1}{(n+1)^2}}=\lim\limits_{n\to\infty}\frac{(n+1)^2}{n^2}=\lim\limits_{n\to\infty}\left(\frac{n+1}{n}\right)^2$$$$\phantom r=\lim\limits_{n\to\infty}\left(1+\frac1n\right)^2=1$$
$$\pink{\text{zu b) }}\quad \sum\limits_{n=1}^\infty b_n\cdot x^n\quad;\quad b_n\coloneqq\frac{n^2}{2^{2n}}=\frac{n^2}{(2^2)^n}=\frac{n^2}{4^n}$$$$r=\lim\limits_{n\to\infty}\left|\frac{b_n}{b_{n+1}}\right|=\lim\limits_{n\to\infty}\frac{\frac{n^2}{4^n}}{\frac{(n+1)^2}{4^{n+1}}}=\lim\limits_{n\to\infty}\frac{4^{n+1}n^2}{4^n(n+1)^2}=4\lim\limits_{n\to\infty}\left(\frac{n}{n+1}\right)^2$$$$\phantom r=4\lim\limits_{n\to\infty}\left(\frac{(n\pink{+1})\pink{-1}}{n+1}\right)^2=4\lim\limits_{n\to\infty}\left(1-\frac{1}{n+1}\right)^2=4$$
$$\pink{\text{zu c) }}\quad \sum\limits_{n=1}^\infty c_n\cdot x^n\quad;\quad c_n\coloneqq\frac{n!}{\sqrt[n]2}=\frac{n!}{2^{\frac1n}}$$$$r=\lim\limits_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right|=\lim\limits_{n\to\infty}\frac{\frac{n!}{2^{\frac1n}}}{\frac{(n+1)!}{2^{\frac{1}{n+1}}}}=\lim\limits_{n\to\infty}\frac{2^{\frac{1}{n+1}}n!}{2^{\frac1n}(n+1)!}=\lim\limits_{n\to\infty}\frac{1}{2^{\frac1n-\frac{1}{n+1}}(n+1)}$$$$\phantom r=\lim\limits_{n\to\infty}\frac{1}{2^{\frac{1}{n^2+n}}(n+1)}<\lim\limits_{n\to\infty}\frac{1}{n+1}=0$$
Die Kleiner-Abschätzung resultuert aus \(2^{\frac{1}{n^2+n}}>1\) bzw. aus \(\frac{1}{2^{\frac{1}{n^2+n}}}<1.\)
