= Solution
Subtract the first four terms from the given full <p-series> sums. Write
$$
S_2=\frac{\pi^2}{6}-\left(1+\frac14+\frac19+\frac1{16}\right)\approx0.2213229557,
$$
and
$$
S_4=\frac{\pi^4}{90}-\left(1+\frac1{16}+\frac1{81}+\frac1{256}\right)\approx0.0035713047.
$$
Solving $I_p(k_p)=S_p$ yields $k_p=5-[(p-1)S_p]^{-1/(p-1)}$. Therefore
$$
\boxed{k_2=5-S_2^{-1}\approx0.482,\qquad k_4=5-(3S_4)^{-1/3}\approx0.464.}
$$
The unrounded shifts are approximately $0.4817158633$ and $0.4638861447$.
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