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?


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 $x>0$.

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.