Question

8: Convergence of summation (n!(n+1)!)/(3n)!

9: Radius of convergence of summation n=0 to infinity (n^3)x^3n/n^4+1

Answer 1

8: Let us study the convergence of summation $\sum_{n=1}^{\infty}\frac{n!(n+1)!}{(3n)!}$. Let us use the D'Alembert's rule.

Taking into account that

$\lim\limits_{n\to \infty}(\frac{(n+1)!(n+2)!}{(3(n+1))!}:\frac{n!(n+1)!}{(3n)!})= \lim\limits_{n\to \infty}(\frac{(n+1)n!(n+2)(n+1)!}{(3n+3)!}\cdot\frac{(3n)!}{n!(n+1)!})= \lim\limits_{n\to \infty}\frac{(n+1)(n+2)}{(3n+3)(3n+2)(3n+1)}= \lim\limits_{n\to \infty}\frac{(1+\frac{1}{n})(1+\frac{2}{n})}{(3+\frac{3}{n})(3+\frac{2}{n})(3n+1)}=0<1$

we conclude that $\sum_{n=1}^{\infty}\frac{n!(n+1)!}{(3n)!}$ is convergent.

9: Let us find the radius of convergence of summation $\sum_{ n=0}^{\infty}\frac{ n^3x^{3n}}{n^4+1}$. Let us use the D'Alembert's rule.

Since $\lim\limits_{n\to \infty}|\frac{ (n+1)^3x^{3(n+1)}}{(n+1)^4+1}:\frac{ n^3x^{3n}}{n^4+1}|= |x|^3\cdot\lim\limits_{n\to \infty}\frac{ (n+1)^3(n^4+1)}{((n+1)^4+1)n^3}= |x|^3\cdot\lim\limits_{n\to \infty}\frac{ (1+\frac{1}{n})^3(1+\frac{1}{n^4})}{(1+\frac{1}{n})^4+\frac{1}{n^4}}=|x|^3\cdot 1=|x|^3<1$imply $|x|<1$, we conclude that the radius $R$ of convergence of summation $\sum_{ n=0}^{\infty}\frac{ n^3x^{3n}}{n^4+1}$ is $R=1.$