I have a problem with solving the convergence of $\sum_{n=1}^{\infty }\frac{(-1)^{[\sqrt{n}]}}{n}$ where $[a]$ means floor function of $a$. I've already made the first thep and rewrote it with the $\sum_{n=1}^{\infty }(-1)^{n}\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{k}$, which converges if and only if the previous one converges. Now I want to use Leibniz criterion, but I cannot prove that the second sum in the previous expression is monotonic. Does anyone know how to do it? (I cannot use Riemann integral, we haven't defined it yet)
convergence of $\sum_{n=1}^{\infty }\frac{(-1)^{[\sqrt{n}]}}{n}$ where $[a]$ means floor function of $a$
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convergence
floor-function
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1[Related](http://math.stackexchange.com/questions/535784/how-find-series-sum-n-1-infty-dfrac-1-sqrtmnna), with a broad interpretation of duplicate it would count as one. – 2017-01-08
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0I would guess that you would have some luck relating $a_n=\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{k}$ to the integral of $\frac{1}{n}$ through left and right Riemann sums. Getting it so that $a_n$ > some definite integral > some other integral > $a_{n+1}$. – 2017-01-08
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Let us recall the definition of the hamronic numbers $$\sum_{k=1}^{N}\frac{1}{k}=H_{N} $$ hence $$S=\sum_{n\geq1}\frac{\left(-1\right)^{\left\lfloor \sqrt{n}\right\rfloor }}{n}=-1+\sum_{n\geq2}\frac{\left(-1\right)^{\left\lfloor \sqrt{n}\right\rfloor }}{n} $$ $$=-1+\sum_{n\geq2}\left(-1\right)^{n}\left(H_{\left(n+1\right)^{2}-1}-H_{n^{2}-1}\right) $$and using $$H_{n}=\log\left(n\right)+\gamma+O\left(\frac{1}{n}\right) $$ we get $$S=-1+\sum_{n\geq2}\left(-1\right)^{n}\log\left(\frac{\left(n+1\right)^{2}-1}{n^{2}-1}\right)+O\left(1\right) $$ $$=-1+\sum_{n\geq2}\left(-1\right)^{n}\log\left(\frac{n\left(n+2\right)}{n^{2}-1}\right)+O\left(1\right)$$ and so we have the convergence for the Leibniz' criterion.