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My first take has been to get rid of the root by doing this:

$$\sum_{n=1}^\infty \frac{\sqrt{n^3+1}}{n^2} =\sum_{n=1}^\infty \sqrt{\frac{n^3+1}{n^4}}$$

Therefore, if $\sum_{n=1}^\infty \frac{n^3+1}{n^4}$ is convergent, the original series must be convergent too. My problem is that once I apply the Cauchy ratio test, I get $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 1$. How can I continue?

PS.: I already know the series is divergent (I checked on Mathematica). What I'm interested in is the process itself, not the result.

5xum
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2 Answers2

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HINT: Use a comparison test. Note that $$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$$ is divergent.

Staki42
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  • Thanks! Another user posted the appropriate test. I don't know why I don't think of these things myself, now it looks so simple :\ –  Jan 26 '17 at 13:25
  • This is just a matter of experience. The first thing you should do when you see a fraction like $\frac{a+b}{c}$ is split it up in $\frac{a}{c} + \frac{b}{c}$. From that it isn't hard to see that this series diverges. :) – Staki42 Jan 26 '17 at 13:34
  • Btw: The series $\sum_{n=1}^{\infty} \frac{1}{n}$ is probably the most famous example for the Cauchy ratio test to fail. That's why it obviously fails for your example too. – Staki42 Jan 26 '17 at 13:36
  • I still rely too much on basic tools (like the ratio test and similar ones). I hope I will get more creative as time passes. Thanks for the hint and the tip :) –  Jan 26 '17 at 13:43
  • You're welcome. You surely will! :) – Staki42 Jan 26 '17 at 14:05
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$$ \frac{\sqrt{n^3+1}}{n^2} \geq \frac{\sqrt{n^3}}{n^2} = \frac{1}{\sqrt{n}}. $$

Siminore
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