$1\mathrm{.}27.$ This problem concerns computing ${\displaystyle \underset{k=1}{\overset{}{}}}\frac{1}{k^2}.$ The exact value of the sum is $\frac{{}^2}{6}.$

$($a$)$ Suppose that the sum is computed using the algorithm given below. Explain how the stopping condition works.

$s=1,ss=0,k=1$

while $sss$

$,k=k+1$

$,ss=s$

$,s=s+\frac{1}{k^2}$

end

$($b$)$ Compute the sum using the procedure in part $($a$)$ and report its value to $12$ digits. At what value of $k$ does the algorithm stop? How many digits of the computed sum are correct?

$($c$)$ Assuming the number of correct digits in your answer is part $($b$)$ is $m,$ compute the sum so the value is correct to$,$ at least, $m+2$ digits. Make sure to explain how you do this. Also, your procedure cannot use the known value of the sum.