Euler discovered and proved that ∑n–2 = π2/6. This is a slowly converging series and the discovery probably required several digits. The naïve summation omits the tail of the series after n terms. That tail sums to about 1/n. Here is how to get a very good estimate of the sum of the tail and thus reduce the error to about 1/(last term)3.
It has to do with approximating n–2 with In = integral of x–2 from n–1/2 to n+1/2.
In = 1/(n – 1/2) – 1/(n + 1/2) = 1/(n2 – 1/4) = n–2/(1 – 1/(4n2)) = n–2∑k≥0(4n2)–k = n–2 + n–2∑k≥1(4n2)–k.
In – n–2 ≃ n–4/4.
∑n≥101 (In – n–2) ≃ ∑n≥101 n–4/4 ≃ 101–3/4.
∑n≥101 (In – n–2) = ∑n≥101 In – ∑n≥101 n–2 ≃ 101–3/4.
but ∑n≥101 In = (Integral x–2 from x=100.5 to ∞) = 1/100.5 . 
#include <stdio.h>
static double sq(double x){return x*x;}
int main(){int i=0; double s = 0;
  double pi = 3.141592653589793238;
  while(i < 100) {++i; s += sq(1./i);}
  s += 1/(i+0.5);
  {double p2 = sq(pi), p2o6 = p2/6;
    printf("%19.16f %19.16f %19.16f %e\n",
      p2, p2o6, s, p2o6-s);}
  return 0;}
This program yields:
9.8696044010893580 1.6449340668482264 1.6449341489411111 -8.209288e-08.
Now I need to evaluate ∑ln(n)n–2 and Dwight gives Integral ln(x)/x2 = –ln(x)/x – 1/x.
Now letting In = integral of ln(x)x–2 from n–1/2 to n+1/2, Mutatis Mutandis (html):
It has to do with approximating ln(n)n–2 with In = integral of ln(x)x–2 from n–1/2 to n+1/2.
In = ln(n – 1/2)/(n – 1/2) – ln(n + 1/2)/(n + 1/2) ≃ ln(n)n–2 + O(n–4)
according to the next small program: 
#include <stdio.h>
#include <math.h>
typedef double f;
static f sq(f x){return x*x;}
static f F(f x){return log(x)/sq(x);}
static f intg(f x){return -log(x)/x - 1/x;}
static f isn(f x){return intg(x + 0.5) - intg(x - 0.5);}
static f err(f x){return F(x) - isn(x);}
int main(){f x = 1;
while (x < 2000000) {printf("%e %e\n", x, err(x)); x *= 10.;}
return 0;}

In – ln(n)n–2 ≃ n–4
∑n≥101 (In – ln(n)n–2) ≃ 101–3
∑n≥101 (In – ln(n)n–2) = ∑n≥101 In – ∑n≥101 ln(n)n–2 ≃ 101–3
but ∑n≥101 In = (Integral ln(x)x–2 from x=100.5 to ∞) = –ln(100.5)/100.5 – 1/100.5 . 
#include <stdio.h>
#include <math.h>
typedef double f;
static f sq(f x){return x*x;}
static f F(f x){return log(x)/sq(x);}
static f intg(f x){return -log(x)/x - 1/x;}
static void mx(int k){int i=0; double s = 0;
  while(i < k) {++i; s += F(i);}
  s -= intg(i+ 0.5);
  printf("%19.16f %d\n", s, k);}

int main(){mx(10); mx(100); mx(1000);
  mx(10000); mx(100000); mx(1000000);
  return 0;}
This program yields: 
 0.9376812399594099 10
 0.9375485917322104 100
 0.9375482548490655 1000
 0.9375482543165695 10000
 0.9375482543158479 100000
 0.9375482543158530 1000000
I conclude that indeed the error is about (last term)3/2 and that the answer to the original problem is:
∑ln(n)n–2 = 0.937548254315853
Perhaps this is ζ'(2).