b

$$\sum _{k=0}^{\infty }\frac{(2k)!}{(k!{)}^k}\cdot (x+2{)}^k=\sum _{k=0}^{\infty }\underset{a_k}{\underbrace{\frac{(2k)!}{(k!{)}^k}}}\cdot (x-\underset{x_0}{\underbrace{(-2)}}{)}^k$$

Bearbeitung §

$$a_n=\frac{(2n)!}{(n!{)}^n}$$

Quotientenmethode §

$$\begin{array}{rl}q& =\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|\\ & =\underset{n\to \infty }{\lim }\left|\frac{\frac{(2n+2)!}{((n+1)!{)}^{n+1}}}{\frac{(2n)!}{(n!{)}^n}}\right|\\ & =\underset{n\to \infty }{\lim }\left|\frac{(2n+2)!}{((n+1)!{)}^{n+1}}\cdot \frac{(n!{)}^n}{(2n)!}\right|\\ & =\underset{n\to \infty }{\lim }\frac{(2n+2)!}{((n+1)!{)}^{n+1}}\cdot \frac{(n!{)}^n}{(2n)!}\\ & =\underset{n\to \infty }{\lim }\frac{(2n)!\cdot (2n+1)\cdot (2n+2)}{(n!{)}^{n+1}\cdot (n+1{)}^{n+1}}\cdot \frac{(n!{)}^n}{(2n)!}\\ & =\underset{n\to \infty }{\lim }\frac{(2n)!\cdot (2n+1)\cdot (2n+2)\cdot (n!{)}^n}{(n!{)}^{n+1}\cdot (n+1{)}^{n+1}\cdot (2n)!}\\ & =\underset{n\to \infty }{\lim }\frac{(2n+1)\cdot (2n+2)}{(n!)\cdot (n+1{)}^{n+1}}\\ & =\underset{n\to \infty }{\lim }\frac{(2n+1)\cdot 2\cdot (n+1)}{(n!)\cdot (n+1{)}^{n+1}}\\ & =\underset{n\to \infty }{\lim }\frac{(2n+1)\cdot 2}{(n!)\cdot (n+1{)}^n}\\ & =\underset{n\to \infty }{\lim }\frac{4\cdot (n+\frac{1}{2})}{(n!)\cdot (n+1{)}^n}\\ & \le \underset{n\to \infty }{\lim }\frac{4\cdot (n+1)}{(n!)\cdot (n+1{)}^n}\\ & =\underset{n\to \infty }{\lim }\frac{4}{(n!)\cdot (n+1{)}^{n-1}}\\ & =0\end{array}$$

$$⟹r=\frac{1}{q}=\frac{1}{0}=\infty$$

Alte Lösung (Falsch)

$$\begin{array}{rl}\dots & \stackrel{\text{L’H}}{=}\underset{n\to \infty }{\lim }\frac{(4n+2{)}^{\prime }}{((n+1{)}^{n-1}{)}^{\prime }}\\ & =\underset{n\to \infty }{\lim }\frac{4}{(n-1)\cdot (n+1{)}^{n-2}\cdot 1}\\ & =\underset{n\to \infty }{\lim }\frac{4}{(n-1)\cdot (n+1)\cdot (n+1{)}^{n-3}}\\ & =\underset{n\to \infty }{\lim }\frac{4}{(n^2-1)\cdot (n+1{)}^{n-3}}\\ & =0\end{array}$$