$($a$)$ Use the Integral Test to show that ${\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{n^4}$ coeverges.

$($b$)$ Find the $10th$ partial sum, $s_{10},$ of the series ${\displaystyle \underset{i=1}{\overset{}{}}}\frac{1}{n^4}.$

$($c$)$ According to the Remainder Estimate for the Integral Test, we know that

${\displaystyle \underset{n1}{\overset{}{}}}\frac{1}{x^4}drx-r_0{\displaystyle \underset{0}{\overset{}{}}}\frac{1}{x^4}dr$

where $s$ is the sum of ${\displaystyle \underset{i=1}{\overset{}{}}}\frac{1}{n^4}$ and $s_n$ is the nth partial ram of ${\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{n^4}.$ Use inequality $(1)$ and $s_{10}$ from part $($b$)$ to find an upper and lower beand for $s.$

$($d$)$ Une the Remainder Batimate for the lenternal Tret to find a valoe of n so that s$,$ is within $0\mathrm{.}00001$ of the sum.