Solution
= Solution
The <shifted integral representation of a p-series tail> evaluates the defining integral as
$$
I_p(k)=\frac1{(p-1)(5-k)^{p-1}},\qquad k<5.
$$
It is strictly increasing in $k$, since $I_p'(k)=(5-k)^{-p}>0$. From the bounds just proved with $a=5$, $I_3(0)<\sum_{n=5}^\infty n^{-3}<I_3(1)$. Since the middle expression equals $I_3(k_3)$, monotonicity proves
$$
\boxed{0<k_3<1.}
$$