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

Question

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!

Answer 1

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, : $$\beta < \dfrac{1}{2} \text{ or } \beta > 2$$

Answer 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.