Bernoulli numbers and $\pi^2$. It is probably well-known that:
$$
\lim_{n\to\infty}\frac{b_{2n}n^2}{b_{2n+2}}=-\pi^2,
$$
where $b_n$ are the Bernoulli numbers.
By a numerical experiment I have found that the quotient
$$
\frac{b_{2n}(n+a_1)(n+a_2)}{b_{2n+2}}
$$
with $a_1=1, a_2=\frac12$ converges to $-\pi^2$ extremely fast. Already $n=35$ gives correctly 20 decimal places of $\pi^2$! No other combination of $a_1$ and $a_2$ is even nearly close to this result. For example $a_1=a_2=\frac12$ gives only two correct decimal places for $n=400$.
What is the secret of this extremely good convergence for $a_1=1, a_2=\frac12$?
 A: Consider the formula (at https://en.wikipedia.org/wiki/Bernoulli_number#Bernoulli_numbers_and_the_Riemann_zeta_function)
$$B_{2n} = {(-1)^{n+1} 2 (2n)! \over (2\pi)^{2n}} \zeta(2n)$$
where $\zeta$ is the Riemann zeta function (https://en.wikipedia.org/wiki/Riemann_zeta_function).
This gives
$${B_{2n} \over B_{2n+2}} = {(-1) (2\pi)^2 \over (2n+1)(2n+2)} {\zeta(2n) \over \zeta(2n+2)} $$
or, multiplying by $(2n+1)(2n+2)/4$,
$$ {B_{2n} (2n+1) (2n+2) \over 4 B_{2n+2}} = -\pi^2 {\zeta(2n) \over \zeta(2n+2)} $$
The left-hand side is the quotient you asked about.  The right-hand side is essentially 1 for large $n$ - more specifically, as $n \to \infty$, we have $\zeta(2n) = 1 + O(1/4^n)$.
So we have
$$ {B_{2n} \over B_{2n+2}} (n+1/2) (n+1) = -\pi^2 (1 + O(1/4^n)). $$
For any other choice of constants $a_1$ and $a_2$, we have
$$ {B_{2n} \over B_{2n+2}} (n+a_1) (n+a_2) = -\pi^2 {(n+a_1)(n+a_2) \over (n+1/2)(n+1)} (1 + O(1/4^n)). $$
The quotient 
$$ {(n+a_1)(n+a_2) \over (n+1/2)(n+1)} = {n^2 + (a_1 + a_2)n + a_1 a_2 \over n^2 + (3/2) n + (1/2) }$$
is $1+O(n^{-2})$ if $a_1 + a_2 = 3/2$ but $\{ a_1, a_2 \} \not = \{ 1/2, 1 \}$, and $1 + O(n^{-1})$ otherwise.  And so we get 
$${B_{2n} \over B_{2n+2}} (n+a_1) (n+a_2) = -\pi^2 (1 + O(1/4^n)) (1 + O(n^{-k})) = -\pi^2 (1 + O(n^{-k})$$
with $k = 1$ or $k = 2$ depending on the choice of $a_1$ and $a_2$.   In particular only $(a_1, a_2) = (1/2, 1)$ or $(a_1, a_2) = (1, 1/2)$ gives the $O(1/4^n)$ error term.
