Proof of convergence of $sum_{n=1}^{infty}frac{(-1)^{lfloor nsqrt{2}rfloor}}{n}$

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 $text{Spec}(alpha)$: $$ text{Spec}(alpha)={lflooralpha nrfloor,:,n=1,2,3,ldots}. $$ Define a function $f(x)$ for positive real $x$: $$ f(x)=sum_{0<n<x}(-1)^{lfloor nsqrt{2}rfloor}. $$ 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 $$ sum_{0<nsqrt{2}<x}(-1)^{lfloor nsqrt{2}rfloor} +sum_{0<n(2+sqrt{2})<x}(-1)^{lfloor n(2+sqrt{2})rfloor} = sum_{0<n< lceil xrceil}(-1)^nin{-1,0}. $$ Since $(-1)^{lfloor n(2+sqrt{2})rfloor}=(-1)^{lfloor nsqrt{2}rfloor}$ we get $$ fleft(frac{x}{sqrt{2}}right)+fleft(frac{x}{2+sqrt{2}}right)in{-1,0}, $$ which leads to inequality: $$ -1le f((1+sqrt{2})x)+f(x)le 0 $$ 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 $sum_{ngeqslant 1}frac {f(n+1)}{n(n+1)}$ 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.