I'm trying to figure out the values of $\beta$ for which $\sum a_n$ converges, where $a_n=\sqrt{1+\dfrac{(-1)^n}{n^\beta}}-1\,$. Here is what I have done:
I tried to re-write $a_n$ as the following, so as to get the monotonic pattern first but I couldn't even forge ahead.
$$a_{2n}=\sqrt{1+\dfrac{1}{(2n)^\beta}}-1\,$$ and $$a_{2n-1}=\sqrt{1+\dfrac{1}{(2n-1)^\beta}}-1\,.$$
Please, can anyone show a way-out?
For $\beta \leq 0$, $a_n \not\to 0$ then the serie does not converge. For $\beta >0$. Consider $f(x) = \sqrt{1+x} -1$. We have $f''(x) = -\frac14 (1+ x)^{-\frac32}$, hence for $|x| \leq 12$ we get $$-C_1=-\frac14 2^{\frac32}\leq f''(-1/2)\leq f''(x) \leq f''(1) = -\frac14 2^{-\frac32} = -C_2.$$ Therefore, for $|x| \leq \frac12$, by Taylor's expansion, we have $$-\frac{C_1}2 {x^2} \leq f(x) -\frac x2 \leq -\frac{C_2}2 {x^2}.$$ For $n$ large enough, we have $(-1)^n/n^\beta \in (-1/2,1/2)$, hence $$\frac{(-1)^n}{2 n^\beta} - \frac{C_1}{2 n^{2\beta}} \leq a_n \leq \frac{(-1)^n}{2 n^\beta} -\frac{C_2}{2 n^{2\beta}},$$ or $$-\frac{(-1)^n}{2 n^\beta} +\frac{C_2}{2 n^{2\beta}}-a_n \leq -\frac{(-1)^n}{2 n^\beta} +\frac{C_1}{2 n^{2\beta}}.$$ Notice that $\sum \frac{(-1)^n}{n^\beta}$ converges for $\beta >0$, hence $\sum a_n$ converges if and only if $\sum n^{-2\beta}$ converges which is equivalent to $\beta > \frac 12$. Hence $\sum a_n$ converges if $\beta > \frac12$.