Can you help me to determine if the limit of this expression is unity?
$$\sum_{n = 2}^{\infty} \frac{ \frac{1}{2n + 1} \cdot \frac{1}{4} \cdot \frac{ 2n}{B_{2n}}}{\sqrt{2}}$$
Where $B_{2n}$ is the Bernoulli Number of $2n$. I have tried this calculation using Wolfram Alpha, but the computation is very expensive, so I have not calculated beyond $n=400$. The values alternate between $y > 1$ and $y < 1$ as $n$ increases, but seem to stick close to $0.995374$ after about 100. This is related to a possible corollary of the Riemann Hypothesis.
Namaste!
No, it's not $1$, I'm pretty sure.
If we can show that the modulus of the terms decreases, then the alternating series test says that the error of the alternating series is at most the first omitted term. Evaluating the summand at $n=10$ yields approximately $-0.000318184$; so the error after $n=9$ is at most that value. The sum after $n=9$ is approximately $-0.9950809$, which is not $0.0003$ away from $-1$.
So we just need it decreasing; it turns out to be decreasing for $n>3$, which you can see heuristically as follows (assuming the approximation holds sufficiently well for some $n \leq 10$). Very strictly I haven't proved this, but I'd be absolutely shocked if it weren't true. The argument above works fine if you can show that $\frac{n}{(2n+1) B_n}$ is decreasing for all $n > M$, some $M \leq 9$.
The Bernoulli numbers grow as $$B_{2n} \sim (-1)^{n-1} 4 \sqrt {\pi n} \left( \frac n {\pi e} \right)^{2n}$$ So your summand is roughly $$\frac{2n}{4 \sqrt{2}} \times \frac{1}{2n+1} \times \frac{(-1)^{n-1} (\pi e)^{2n}}{4 \times n^{2n} \sqrt{\pi n}}$$ which is in modulus $$\frac{2n}{4 \sqrt{2}} \times \frac{1}{2n+1} \times \frac{(\pi e)^{2n}}{4 \times n^{2n} \sqrt{\pi n}} \leq \frac{1}{16 \sqrt{2}} \times \frac{(\pi e)^{2n}}{n^{2n} \sqrt{\pi n}}$$
This is decreasing in $n$ for $n>3$.