= Solution
Use the stipulated interpolation $k_3\approx(k_2+k_4)/2$. The unrounded values give $k_3\approx0.4728010040$, and hence
$$
\sum_{n=1}^\infty n^{-3}\approx1+\frac18+\frac1{27}+\frac1{64}+\frac1{2(5-0.4728010040)^2}\approx1.2020575995.
$$
Thus a sensible reported estimate is
$$
\boxed{\sum_{n=1}^\infty n^{-3}\approx1.20206.}
$$
Using the rounded shifts from part (ii) instead gives $k_3\approx0.473$ and a sum about $1.202059745$, with the same six-significant-figure estimate. The interpolation of shifts is an assumption of this approximation; the <integral test> bounds alone do not prove that many digits of accuracy.
Back to article page
