# Convergence of $\sum_{n=1}^\infty \frac{1}{(n^{2/\beta}(n^{2\beta}+1))^{1/5}}$

I'm stuck on a problem about $$\sum_{n=1}^\infty \frac{1}{(n^{\frac{2}{\beta}}(n^{2\beta}+1))^{\frac{1}{5}}}$$. I need to find for which values of $$\beta$$ this series converge (the correct answer is $$0 < \beta < \frac{1}{2}$$) but no matter how I manipulate it, I don't know how to isolate beta.

I've tried Leibniz criterium and simplify it as $$n^{\frac{2}{\beta}}(n^{2\beta}+1) \to \inf$$ but when I do that, I find that $$\beta$$ must be between $$0 < \beta < 2$$ which is false obviously. I'm kind of lost, at first I thought it was because it's a polynomial but since we are looking at the limit it doesn't make sense...

I don't see any others criterium that could help so I tried to compare it to a series I know but I find it hard to put this series in a form I could find one. I think I need to isolate $$\beta$$ at some point, but I can't? Thanks for reading!

Notice that : $$\frac{1}{(n^{\frac{2}{\beta}}(n^{2\beta}+1))^{\frac{1}{5}}} \sim \frac{1}{n^{\frac{2}{5 \beta} + \frac{2\beta}{5}}}$$ We deduce that the series : $$\sum \frac{1}{(n^{\frac{2}{\beta}}(n^{2\beta}+1))^{\frac{1}{5}}}$$ converges if, and only if, : $$\frac{2}{5 \beta} + \frac{2\beta}{5} > 1$$ if, and only if, : $$2 \beta^2 - 5 \beta + 2 > 0$$ if, and only if, : $$(2 \beta - 1) (\beta - 2) > 0$$ if, and only if, : $$\beta < \dfrac{1}{2} \text{ or } \beta > 2$$

Clearly, the series diverges for $$\beta < 0$$ since as $$n \to \infty$$, we have

$$a_n =\frac{1}{[n^{\frac{2}{\beta}}(n^{2\beta} +1)]^{\frac{1}{5}}}= \frac{n ^{\frac{2}{5|\beta|}}}{(n^{-2|\beta|}+1)^{\frac{1}{5}}}\to +\infty$$

We also have

$$a_n =\frac{1}{[n^{\frac{2}{\beta}}(n^{2\beta} +1)]^{\frac{1}{5}}}< b_n = \frac{1}{n^{\frac{2}{5\beta}+\frac{2\beta}{5}}},$$

and $$\frac{2}{5\beta}+\frac{2\beta}{5} >1$$ if $$p(\beta)=2\beta^2 - 5\beta +2 >0$$. The roots of the quadratic polynomial $$p$$ are $$\frac{1}{2}$$ and $$2$$ and it is easy to check that $$p(\beta) >0$$ when $$\beta< \frac{1}{2}$$ and $$\beta > 2$$. Hence, $$\sum b_n$$ is a convergent $$p$$-series and the series in question $$\sum a_n$$ converges by the comparison test for $$0<\beta < \frac{1}{2}$$ and $$\beta > 2$$.

For any $$\beta > 0$$ and all sufficiently large $$n$$, we have $$n^{2\beta}>1$$ and $$a_n =\frac{1}{[n^{\frac{2}{\beta}}(n^{2\beta} +1)]^{\frac{1}{5}}}> b_n = \frac{1}{[n^{\frac{2}{\beta}}(2n^{2\beta})]^{\frac{1}{5}}} =\frac{1}{2^{\frac{1}{5}}n^{\frac{2}{5\beta}+\frac{2\beta}{5}}}$$

Inspection of $$p(\beta)$$ from the previous analysis shows that $$p(\beta) \leqslant 0$$ and $$n^{2\beta}\leqslant 1$$ for $$\frac{1}{2} \leqslant \beta \leqslant 2$$. Hence, $$\sum b_n$$ is a divergent $$p$$-series and, by comparison, the series $$\sum a_n$$ diverges for $$\beta$$ in this range.