convergence range of $\sum_{n=1}^\infty\frac{(-1)^{[\sqrt n]}}{n}x^n$

I need help finding the range of convergence of the power series:

$$\sum_{n=1}^{\infty}\frac{(-1)^{[\sqrt n]}}{n}x^n$$

I found the radius of convergence $R=\frac{1}{\sqrt[n]{|a_n|}}=1$

my problem is with checking the edges $1$ and $-1$

I tried using a few tests but I don't know how to deal with the floor function.

any suggestions?

• Are you looking for a series or a sequence? You have written down a sequence. – Jason Jul 13 '17 at 14:34
• @Jason thank you, fixed – segevp Jul 13 '17 at 14:36

We wish to determine whether the series $\sum_{n=1}^{\infty}\frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{n}$ exists.

We start by writing

\begin{align} \sum_{n=1}^{(L+1)^2-1} \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{n}&=\sum_{k=1}^L \sum_{n=k^2}^{(k+1)^2-1}\frac{(-1)^k}{n}\\\\ &=\sum_{k=1}^L (-1)^k \sum_{n=k^2}^{(k+1)^2-1}\frac{1}{n}\\\\ \end{align}

From the Euler-Maclaurin Summation Formula, we have

\begin{align} \sum_{n=k^2}^{(k+1)^2-1}\frac{1}{n}&=\int_{k^2-1}^{(k+1)^2-1}\frac1x\,dx+O\left(\frac{1}{k^2}\right)\\\\ &=\log\left(\frac{(k+1)^2-1}{k^2-1}\right)+O\left(\frac1{k^2}\right)\\\\ &=\frac2k +O\left(\frac1{k^2}\right) \end{align}

Inasmuch as $\sum_{k=1}^\infty \frac{(-1)^k}{k}$ converges, then the series of interest converges likewise.

• Not sure why you introduce $f(x).$ For one thing, what does $f(1)$ even mean until you prove it has meaning? – zhw. Jul 13 '17 at 17:21
• @zhw. I've edited. – Mark Viola Jul 13 '17 at 17:45