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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 $a_1=1, a_2=\frac12$?

user
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  • For large n the addition of the constants a1 and a2 really doesn't make a difference. I mean n of order 100 – lux Apr 09 '19 at 13:50
  • @lux: it doesn’t affect the limit but it can affect the rate of convergence; that is what is being asked about here. – Clayton Apr 09 '19 at 13:54
  • @lux It could be expected that it does not make difference. The astonishing fact is that the difference is huge. Try it yourself. – user Apr 09 '19 at 14:00
  • @Clayton, I agree with your comment, although after n gets sufficiently large we don't really care. To the op, just to clarify, there is no doubt it makes no difference whatsoever to the limit (I. E. You will get - pi^2 for n sufficiently large) but perhaps for small n you get some answers that are closer to the limit than without the constants. – lux Apr 09 '19 at 14:08
  • @lux: I never had doubts about the limit. But you can convince yourself by trying that with a different choice of constants you never get such a precision for "$n$ of order 100". Even with $n=1000000$ you will get only 5 correct decimal places. – user Apr 09 '19 at 14:17
  • $\lim_{n \to \infty}\frac{B_{2n+2j-2}}{B_{2n+2j}}\cdot (n+j)\left(n+j-\frac{1}{2}\right)=-\pi^2$. For higher of values of $j$, it converges quicker. – Endgame Apr 16 '19 at 20:38
  • @Endgame It is exactly the same with my $n$ equal to your $n+j-1$. – user Apr 16 '19 at 20:41

1 Answers1

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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.

Michael Lugo
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