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$\sum_{n=1}^\infty{(\sqrt{n}-2)^2 \over n^2+\sqrt{n^4+1} }$

I tried using the ratio test but i cant cancel anything out.

2 Answers 2

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Limit comparison test:

$$\lim_{n\to\infty}\frac{\frac{(\sqrt n-2)^2}{n^2+\sqrt{n^4+1}}}{\frac1{2n}}=1$$

Thus, as $k\to\infty$,

$$\sum_{n=1}^k{(\sqrt{n}-2)^2 \over n^2+\sqrt{n^4+1} }\sim\sum_{n=1}^k\frac1{2n}$$

which is the harmonic series and diverges.

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    The limit is $\frac12$. – 2017-01-22
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    @user296113 Sorry, that's right. (honestly just compared the powers, but I should've noticed that, thanks) – 2017-01-22
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    On the last line, it might be better *not* to start the index at $n=1$. – 2017-01-22
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    @OlivierOloa Why is that? – 2017-01-22
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    The $\sim$ is true when $n \to \infty$. – 2017-01-22
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    @OlivierOloa Like so? – 2017-01-22
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    If the limit was $1$, we wouldn't be able to conclude anything. Just a friendly reminder. – 2017-01-22
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    The last line needn't be true. Since $\sum \frac 1n$ diverges, you get asymptotic equivalence on the partial sums, not the tails. – 2017-01-22
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    @LeGrandDODOM Actually I think it's the tails. Last time I checked the proof of the limit comparison test, that is. – 2017-01-22
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    @SimplyBeautifulArt No, with **divergent non negative** series, you get equivalence of [partial sums](http://imgur.com/ONyeBRH) – 2017-01-22
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    @LeGrandDODOM Huh, ok. Never knew... – 2017-01-22
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By equivalence: $$(\sqrt n-2)^2\sim_\infty\sqrt n^2=n,\enspace n^2+\sqrt{n^4+1}\sim_\infty 2n^2, \enspace\text{whence}\quad\frac{(\sqrt n-2)^2}{n^2+\sqrt{n^4+1}}\sim_\infty \frac{n}{2n^2}=\frac1{2n},$$ which diverges.

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    :D Which was my line of logic when I wrote my answer. – 2017-01-22