$(******)$ We would like to evaluate the series ${\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{(\sqrt[\phantom{2}]{36+n^2}{)}^3}.$

$($a$)$ Use the error bound from Problem $22$ of this assignment to bound the error incurred using ${\displaystyle \underset{n=1}{\overset{10}{}}}\frac{1}{(\sqrt[\phantom{2}]{36+n^2}{)}^3}$ approximate the entire series.

${\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{(\sqrt[\phantom{2}]{36+n^2}{)}^3}-{\displaystyle \underset{n=1}{\overset{10}{}}}\frac{1}{(\sqrt[\phantom{2}]{36+n^2}{)}^3}$

$($b$)$ Using a computer, we find ${\displaystyle \underset{n=1}{\overset{10}{}}}\frac{1}{(\sqrt[\phantom{2}]{36+n^2}{)}^3}$~~$0\mathrm{.}626452279375411.$ Given this information, give the smallest possible interval that

contains the value ${\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{(\sqrt[\phantom{2}]{36+n^2}{)}^3}.$ Include at least $6$ decimal places in your answers.

${\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{(\sqrt[\phantom{2}]{36+n^2}{)}^3}$