= Zeta series truncation error
{title2=$R_N(s)=\zeta(s)-\sum_{n=1}^{N}n^{-s}$}
For a real <exponent> $s>1$ and a <positive integer> $N$, the <series truncation error> in the defining <Riemann zeta function> <series> is
$$
R_N(s)=\sum_{n=N+1}^{\infty}n^{-s}>0.
$$
The <integral comparison> for the positive decreasing <function> $x^{-s}$ gives the explicit bounds
$$
\boxed{\frac{(N+1)^{1-s}}{s-1}\leq R_N(s)\leq\frac{N^{1-s}}{s-1}}.
$$
On each unit <interval>, the area under $x^{-s}$ lies between its endpoint heights. Summing from the first omitted term gives the lower <integral> from $N+1$ to infinity and the upper <integral> from $N$ to infinity. This uses positivity and monotonicity, not an assumption that the last computed term equals the remaining error.
At $N=10$, the bounds are $1/11\leq R_{10}(2)\leq1/10$, whereas $1/(3\cdot11^3)\leq R_{10}(4)\leq1/(3\cdot10^3)$. The errors above fit both intervals. Merely looking at the eleventh term would give $1/121\approx0.008264$ for squares, far smaller than the actual remaining $0.095166$.
\Image[/zeta-even-power-tail-bounds.png]
{title=Integral bounds enclose the full remaining error of two zeta series}
{description=Original computed errors, obtained by subtracting finite <partial sums> from the exact even zeta values. Shaded bands show the upper and lower <integral comparison> bounds. The logarithmic axes reveal decay proportional to $1/N$ for squares and $1/N^3$ for fourth powers. These are mathematical computations, not measured data.}
For a chosen absolute error target $\varepsilon>0$, the sufficient condition is $N\geq((s-1)\varepsilon)^{-1/(s-1)}$, rounded upward to an <integer>. Thus $\varepsilon=10^{-4}$ requires $N=10000$ by this bound for squares but only $N=15$ for fourth powers. This is a guarantee from an upper bound, not a claim that those are the smallest successful term counts.
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