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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 $x \in \mathbb R^+_0$?

Is this the correct way to go about this or am I overlooking something?

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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.

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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 $x>0$.

Edit: After I posted my answer I saw that @RRL had posted a more complete answer.

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  • $\begingroup$ That was a typo. Yes, it is basically the same answer. $\endgroup$ – mlerma54 Dec 3 '18 at 20:44
  • $\begingroup$ I didn't see your answer until I posted mine, otherwise I would have not posted it. Yours is also more complete since it discusses non uniform convergence. $\endgroup$ – mlerma54 Dec 3 '18 at 20:47
  • $\begingroup$ No problem then, $\endgroup$ – RRL Dec 3 '18 at 20:47

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