TAOCP 1.2.7 Exercise 12
By the definition of generalized harmonic numbers, H_{\infty}^{(1000)}=\sum_{k=1}^{\infty}\frac1{k^{1000}}=\zeta(1000).
Section 1.2.7: Harmonic Numbers
Exercise 12. [M10] Evaluate $H_{\infty}^{(1000)}$ correct to at least 100 decimal places.
Verified: no
Solve time: -
By the definition of generalized harmonic numbers,
$$ H_{\infty}^{(1000)}=\sum_{k=1}^{\infty}\frac1{k^{1000}}=\zeta(1000). $$
For even integers, Euler's formula gives
$$ \zeta(2m)=\frac{(-1)^{m+1}B_{2m}(2\pi)^{2m}}{2(2m)!}. $$
In the present case, direct numerical evaluation is simpler. Since
$$ \sum_{k=2}^{\infty}\frac1{k^{1000}} <\sum_{k=2}^{\infty}\frac1{2^{1000}}\frac1{2^{k-2}} =\frac1{2^{999}}, $$
the tail beyond the first term is less than $10^{-300}$. Hence the decimal expansion agrees with $1$ through far more than $100$ decimal places:
$$ H_{\infty}^{(1000)} =1.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\cdots $$
Therefore,
$$ \boxed{H_{\infty}^{(1000)}=1.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000} $$
correct to at least $100$ decimal places.