$(4pts)9{\displaystyle \underset{n=1}{\overset{}{}}}{n}^2(x-5{)}^n$

$4,6\underset{n}{lim}|\frac{a_{n+1}}{a_n}|<mn>1</mn>$

$\underset{n}{lim}|\frac{(n+1{)}^2(x-5{)}^{n+1}}{n^2(x-5{)}^n}|<mn>1</mn>$

$\underset{n}{lim}|(x-5)\frac{(n+1{)}^2}{n^2}|<mn>1</mn>$

$|x-5|\underset{n}{lim}(\frac{n+1}{n}{)}^2<mn>1</mn>$

$|x-5|*1^2<mn>1</mn>$

$|x-5|<mn>1</mn>$

$-1$

$So$ the interval $of$ convergence $is4,6$