How the prove that the Dirichlet series converges?

Question

I am ready to calculate an alternated series of the type of Dirichlet $L$-function. The series, as it is, is $$ L = {\sum_{n=1,~n\neq0(\textrm{mod}~3)}^{\infty}}\frac{(-1)^{n+1}\cdot\color{red}{(-1)^{[n/3]}}}{\sqrt{n}} $$ where the symbol $[x]$ denotes the integral part of $x$, or equivalently $$ L=1-\frac{1}{\sqrt2}+\frac{1}{\sqrt4}-\frac{1}{\sqrt5}+\frac{1}{\sqrt7}-\frac{1}{\sqrt8}+\dots~. $$ Here I just want to ask two questions. Firstly, how can we prove that the series converges(The series tends to be 0.48 approximatly). Secondly, can we transform it to a rapid series so that we can work it out quickly?

Answer 1

We have \begin{align} \sum_{n \neq 0\pmod3}^{3k+2}\dfrac{(-1)^{n+1}(-1)^{\lfloor n/3 \rfloor}}{\sqrt{n}} & = \sum_{n=0}^{k} \left(\dfrac{(-1)^{3n+2}(-1)^{\lfloor (3n+1)/3 \rfloor}}{\sqrt{3n+1}} + \dfrac{(-1)^{3n+3}(-1)^{\lfloor (3n+2)/3 \rfloor}}{\sqrt{3n+2}}\right)\\ & = \sum_{n=0}^k \left(\dfrac1{\sqrt{3n+1}} - \dfrac1{\sqrt{3n+2}}\right) = \sum_{n=0}^k \left(\dfrac{\sqrt{3n+2} - \sqrt{3n+1}}{\sqrt{3n+1}\sqrt{3n+2}}\right)\\ & = \sum_{n=0}^k \dfrac1{\sqrt{3n+1}\sqrt{3n+2}\left(\sqrt{3n+1}+\sqrt{3n+2} \right)}\\ & < \dfrac1{\sqrt2(1+\sqrt2)} + \sum_{n=1}^k \dfrac1{2(3n)^{3/2}} = \dfrac1{\sqrt2(1+\sqrt2)} + \dfrac1{6\sqrt3} \sum_{n=1}^k \dfrac1{n^{3/2}} \end{align} which clearly converges.