Series acceleration and zeta(2)
(This is Yan Sheng.) Today's post started with this question: IMC 2015 Q6 : Prove that $\displaystyle\sum_{n=1}^\infty\frac1{\sqrt n(n+1)}<2$. The interested reader can pause here to try it out, while I helpfully fill the next few sentences with flavour text to delay spoilers. One of my favourite hobbies is to cosplay as Euler. No, not literally (though I'm sure some of you freaks who read this blog would love to see it), but by trying to manipulate infinite series in all sorts of fun ways. I've already written here before about hand computation and zeta(2) , and this post will be more of the same mucking around. Theodorus's constant For today's problem, seeing that the terms are of order $n^{-3/2}$, the first instinct should be to compare with a telescoping series whose tail sums are of order $n^{-1/2}$. Indeed, it's completely routine to check that$$2\left(\frac1{\sqrt n}-\frac1{\sqrt{n+1}}\right)-\frac1{\sqrt n(n+1)}=\frac{(\sqrt{n+1}-\sqrt n)^2}...