Mình cần giúp ạ. Cảm ơn

\(\begin{array}{l}
a)\quad \displaystyle\sum\limits_{n=3}^{\infty}\dfrac{7^{3n}}{(2n-5)!}\\
\text{Đặt}\ u_n = \dfrac{7^{3n}}{(2n-5)!},\ \text{ta có:}\\
\quad \lim\limits_{n\to \infty}\dfrac{u_{n+1}}{u_n}\\
 = \lim\limits_{n\to \infty}\dfrac{7^3(2n-5)!}{(2n-3)!}\\
 = \lim\limits_{n\to \infty}\dfrac{343}{(2n-3)(2n-4)}\\
= 0 < 1\\
\text{Vậy theo tiêu chuẩn D'Alembert, chuỗi đã cho hội tụ}\\
b)\quad \displaystyle\sum\limits_{n=1}^{\infty}\dfrac{(n!)^2}{(2n)!}\\
\text{Đặt}\ u_n = \dfrac{(n!)^2}{(2n)!},\ \text{ta có:}\\
\quad \lim\limits_{n\to \infty}\dfrac{u_{n+1}}{u_n}\\
=\lim\limits_{n\to \infty}\dfrac{[(n+1)!]^2.(2n)!}{(2n+2)!.(n!)^2}\\
=\lim\limits_{n\to \infty}\dfrac{(n+1)^2}{(2n+2)(2n+1)}\\
= \dfrac14 < 1\\
\text{Vậy theo tiêu chuẩn D'Alembert, chuỗi đã cho hội tụ}\\
c)\quad \displaystyle\sum\limits_{n=1}^{\infty}\left(\dfrac{n}{3n-1}\right)^{2n-1}\\
\text{Đặt}\ u_n = \left(\dfrac{n}{3n-1}\right)^{2n-1},\ \text{ta có:}\\
\quad \lim\limits_{n\to \infty}\sqrt[n]{u_n}\\
= \lim\limits_{n\to \infty}\left(\dfrac{n}{3n-1}\right)^{\dfrac{2n-1}{n}}\\
= \lim\limits_{n\to \infty}e^{\displaystyle{\ln\left(\dfrac{n}{3n-1}\right)^{\dfrac{2n-1}{n}}}}\\
= e^{\displaystyle{\lim\limits_{n\to \infty}\left[\dfrac{2n-1}{n}\cdot \ln\left(\dfrac{n}{3n-1}\right)\right]}}\\
= e^{\displaystyle{2\ln\dfrac13}}\\
= \dfrac19<1\\
\text{Vậy theo tiêu chuẩn Cauchy, chuỗi đã cho hội tụ}\\
d)\quad \displaystyle\sum\limits_{n=1}^{\infty}\dfrac{3}{3^n}\left(\dfrac{n+1}{n}\right)^{n^2}\\
\text{Đặt}\ u_n = \dfrac{3}{3^n}\left(\dfrac{n+1}{n}\right)^{n^2},\ \text{ta có:}\\
\quad \lim\limits_{n\to \infty}\sqrt[n]{u_n}\\
= \lim\limits_{n\to \infty}\dfrac{\sqrt[n]{3}}{3}\left(\dfrac{n+1}{n}\right)^{n}\\
= \dfrac13\cdot\lim\limits_{n\to \infty}3^{\tfrac 1n}\cdot \lim\limits_{n\to \infty}e^{\displaystyle{\ln\left(\dfrac{n+1}{n}\right)^{n}}}\\
= \dfrac13\cdot e^{\displaystyle{\lim\limits_{n\to \infty}\left[n\cdot \ln\left(\dfrac{n+1}{n}\right)\right]}}\\
= \dfrac13\cdot e^{\displaystyle{\lim\limits_{n\to \infty}\dfrac{\ln\left(\dfrac{n+1}{n}\right)}{\dfrac1n}}}\\
= \dfrac13\cdot e^{\displaystyle{\lim\limits_{n\to \infty}\dfrac{n}{n+1}}}\\
= \dfrac{e}{3} < 1\\
\text{Vậy theo tiêu chuẩn Cauchy, chuỗi đã cho hội tụ}\\
\end{array}\)