# Can we show that $\sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt n \rfloor}}{n}$ converges by Dirichlet's test?

by JustAnAmateur   Last Updated August 01, 2020 14:20 PM

I know that this is a well-known series, but I didn't fiind a proof for its convergence using Dirichlet's test and I think that this can be done. So, can we show that $$\sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt n \rfloor}}{n}$$ converges by Dirichlet's test?
My attempt: For $$n \in \mathbb{N}$$ we define $$x_n=\frac{1}{n}$$ and $$y_n=(-1)^{\lfloor \sqrt n \rfloor}$$. It is clear that the sequence $$(x_n)_{n\in \mathbb{N}}$$ is a decreasing sequence that converges to $$0$$. We only need to show that the sequence $$(s_n)_{n\in \mathbb{N}}$$, $$s_n=\sum\limits_{k=1}^n y_k$$ is bounded.
We have that $$s_{n^2}=\sum\limits_{k=1}^n\left(\sum\limits_{i=k}^{k^2+2k}(-1)^i\right)$$, $$\forall n\in \mathbb{N}$$ and now I am stuck. To me, it should be pretty obvious that this is bounded, but I don't know, I may be wrong and then Dirichlet's test may not be applied.

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