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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.

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    Where do you get $\sum t_n$ CV $\implies\sum\sqrt{t_n}$ CV from ? This is wrong. – 2017-01-26
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    I don't understand your question. – 2017-01-26

2 Answers 2

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

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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 :\ – 2017-01-26
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    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. :) – 2017-01-26
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    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. – 2017-01-26
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    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 :) – 2017-01-26
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    You're welcome. You surely will! :) – 2017-01-26
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$$ \frac{\sqrt{n^3+1}}{n^2} \geq \frac{\sqrt{n^3}}{n^2} = \frac{1}{\sqrt{n}}. $$