Proof of convergence of $\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor n\sqrt{2}\rfloor}}{n}$ I had this task long long time ago in a calculus class. I remember it had a very elegant solution using nice property of $\text{Spec}(\sqrt{2})$,
where
$$
\text{Spec}(\alpha)=\{\lfloor\alpha n\rfloor\,:\,n=1,2,3,\ldots\}.
$$
I wasn't able to recreate the solution so I asked this question here.
Thanks to zwim I was able to recall the original solution. See my answer below.
 A: Thanks to zwim I was able to recreate original solution which was beautiful. I just wanted recall that solution because it uses nice property from

Graham, Ronald L.; Knuth, Donald E.; Patashnik, Oren, Concrete mathematics: a foundation for computer science., Amsterdam: Addison-Wesley Publishing Group. xiii, 657 p. (1994). ZBL0836.00001, Section 3.2, Page 77.

Define a multiset of integers $\text{Spec}(\alpha)$:
$$
\text{Spec}(\alpha)=\{\lfloor\alpha n\rfloor\,:\,n=1,2,3,\ldots\}.
$$
Define a function $f(x)$ for positive real $x$:
$$
f(x)=\sum_{0<n<x}(-1)^{\lfloor n\sqrt{2}\rfloor}.
$$
There is a beautiful proof in "Concrete Mathematics" that $\text{Spec}(\sqrt{2})$ and $\text{Spec}(2+\sqrt{2})$ form a disjoint partition of the set of positive integers, thus
$$
\sum_{0<n\sqrt{2}<x}(-1)^{\lfloor n\sqrt{2}\rfloor}
+\sum_{0<n(2+\sqrt{2})<x}(-1)^{\lfloor n(2+\sqrt{2})\rfloor}
=
\sum_{0<n< \lceil x\rceil}(-1)^n\in\{-1,0\}.
$$
Since $(-1)^{\lfloor n(2+\sqrt{2})\rfloor}=(-1)^{\lfloor n\sqrt{2}\rfloor}$ we get
$$
f\left(\frac{x}{\sqrt{2}}\right)+f\left(\frac{x}{2+\sqrt{2}}\right)\in\{-1,0\},
$$
which leads to inequality:
$$
-1\le f((1+\sqrt{2})x)+f(x)\le 0
$$
for all positive real $x$.
Now, by induction over $k=0,1,2,\ldots$ one can prove such claim:
$$
0<x<(1+\sqrt{2})^k \implies |f(x)|\le k.
$$
As mentioned in https://artofproblemsolving.com/community/c7h39398p245740 (link provided by zwim)
using the Abel transform we conclude that our series converges if and only if the series $\sum_{n\geqslant 1}\frac {f(n+1)}{n(n+1)}$ converges and, in case of convergence, they converge to the same sum.
It is enough to show that $|f(x)|=O(\log x)$ - which we did.
