# Series convergence with parameter:

Determine whether the following series : $$\sum_{n=2}^\infty \frac {1+xn}{\sqrt{n^2+n^6x}}, x \in \mathbb R^+_0$$ converges absolutely, conditionally or diverges.

I tried to estimate the series for $$n > x$$ using the following: $$\sum_{n=2}^\infty \frac {1+xn}{\sqrt{n^2+n^6x}} \leq \sum_{n=2}^\infty \frac {2xn}{\sqrt{n^6x}} = 2\sqrt x\sum_{n=2}^\infty n^{-2}$$

Which would mean that the series is absolutely convergent for every $$x \in \mathbb R^+_0$$?

Note that the series diverges if $$x = 0$$.

You need a slight correction to your argument. For fixed $$x > 0$$, we have for all $$n > 1/x$$,

$$\frac{1+xn}{\sqrt{n^2 + n^6x}} \leqslant \frac{2nx}{n^3\sqrt{x}} = \frac{2\sqrt{x}}{n^2},$$

and by the comparison test we have pointwise convergence for all $$x > 0$$.

The convergence is not uniform for $$x \in (0,\infty)$$. Note that

$$\sup_{x \in [0,\infty)}\sum_{n=m+1}^{\infty} \frac{1+xn}{\sqrt{n^2 + n^6x}} >\sup_{x \in [0,\infty)}\sum_{n=m+1}^{2m} \frac{1}{n\sqrt{1 + n^4x}} > \sup_{x \in [0,\infty)}\frac{m}{(2m)\sqrt{1+(2m)^4x}} \\ = \sup_{x \in [0,\infty)}\frac{1}{2\sqrt{1+16m^4x}} \geqslant \underbrace{\frac{1}{2\sqrt{1+16m^4{m^{-4}}}}}_{\text{ taking } x = m^{-4}} = \frac{1}{2\sqrt{17}}$$

and the RHS does not converge to $$0$$ as $$m \to \infty$$.

The convergence is uniform on any compact interval $$[a,b]$$ with $$a > 0$$. I will leave this for you to prove.

Perhaps you meant $$n > \frac{1}{x}$$? Then $$\sum_{n=2}^{\infty} \frac{1+xn}{\sqrt{n^2 + n^6 x}} \leq \sum_{n=2}^{\infty} \frac{2nx}{\sqrt{n^6 x}} = \sum_{n=2}^{\infty} \frac{2 \sqrt{x}}{n^2} = 2\sqrt{x} \sum_{n=2}^{\infty} \frac{1}{n^2}$$ which is convergent for each $$x>0$$.