How to prove the convergence of $\sum\limits_{n=1}^{+\infty}\frac{(-1)^{⌊\sqrt{n}⌋}}{n^p}$ when 1/2How should I prove the convergence of $\sum\limits_{n=1}^{+\infty}\frac{(-1)^{⌊\sqrt{n}⌋}}{n^p}$ when 1/2<p<1? I have found that the sum of the original series is equal to $\sum\limits_{n=1}^{+\infty}(-1)^n\sum\limits_{i=n^2}^{n^2+2n}\frac{1}{i^p}$. I'm trying to use Derichlet test, so the question is transformed into: How to prove that $\sum\limits_{i=n^2}^{n^2+2n}\frac{1}{i^p}$ is monotonically decreasing to 0 when 1/2<p<1.
I have used integrals to estimate that $\int_{n^{2}}^{n^{2}+2 n+1} \frac{1}{x^{p}} d x \leqslant \sum\limits_{i=n^{2}}^{n^{2}+2 n} \frac{1}{i^p} \leqslant \int_{n^{2}-1}^{n^{2}+2 n} \frac{1}{x^{p}} d x$ and I can prove that the limit of $ \sum\limits_{i=n^{2}}^{n^{2}+2 n} \frac{1}{i^p} (n\rightarrow+\infty)$ is 0 since $ \sum\limits_{i=n^{2}}^{n^{2}+2 n} \frac{1}{i^p} \leqslant \frac{2n+1}{n^{2p}}$. But I find it difficult to prove that it is monotonically decreasing. Shall I switch to another method? Thanks for your help.
 A: Indeed $\sum_{n=1}^\infty(-1)^{\lfloor\sqrt{n}\rfloor}n^{-p}=\sum_{n=1}^\infty(-1)^n s_n$ where $s_n=\sum_{k=n^2}^{n^2+2n}k^{-p}$, and we have to prove that the last series converges (for $1/2<p<1$). But, instead of trying to show that $n\mapsto s_n$ decreases, one might just put $\color{blue}{s_n=2n^{1-2p}+a_n}$ and prove that $\sum_{n=1}^\infty|a_n|$ converges (then $\sum_{n=1}^\infty(-1)^n s_n$ is the sum of two series: $2\sum_{n=1}^\infty(-1)^n n^{1-2p}$ which converges by the alternating series test, and $\sum_{n=1}^\infty(-1)^n a_n$ which converges absolutely).
We use $(1+x)^c=1+cx+O(x^2)$ as $x\to 0$. At $c=1-p$ and $x=1/k$, we get $$\frac{(k+1)^{1-p}-k^{1-p}}{1-p}=k^{-p}+O(k^{-p-1})\qquad\text{as }k\to\infty$$ which, after summing over $k$ (with $n^2\leqslant k\leqslant n^2+2n$), gives $$\frac{(n+1)^{2-2p}-n^{2-2p}}{1-p}=s_n+O(n\cdot n^{-2p-2})\qquad\text{as }n\to\infty$$ but the LHS is $2n^{1-2p}+O(n^{-2p})$, by the above at $c=2-2p$ and $x=1/n$.
Hence $a_n=O(n^{-2p})$, which is sufficient.
