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I am unsure if my process for finding the comparison series is correct. Is it?

Does this series converge or diverge?

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

so I'm not sure if there's a more formal way to do this but I picked $\frac{1}{n}$ as the comparison series. I took the square root of $n^4$ which is $n^2$ and left it in the numerator. then I figured $\frac{n^2}{n^3} = \frac{1}{n}$. Is there a better way to do this?

Then I did a Limit Comparison test:

$$\frac{(n^4+1)^\frac{1}{2}}{n^3+n} * n = \frac{(n^6 + n^2)^{\frac{1}{2}}}{n^3 + n} = \frac{n^3 + n}{n^3 + n} = 1$$

So the original series diverges too. Is this correct?

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    $\begingroup$ $(n^6 + n^2)^{1/2} \ne n^3 + n$. $\endgroup$ – Daniel Schepler Apr 9 at 23:32
  • $\begingroup$ Try to prove that $\frac{\sqrt{n^4 + 1}}{n^3+n}\ge\frac1{n+1}$ (square both sides and cross-multiply) $\endgroup$ – robjohn Apr 11 at 4:48
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Hint: $\dfrac{\sqrt{n^4+1}}{n^3+n} > \dfrac{1}{2n}$

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Your approach is good but the equalities you have stated are obviously false. To find the limit of $\frac {{(n^{4}+1)^{1/2} n}} {n^{3}+n}$ write this as $\frac {(n^{4}+1)^{1/2}} {n^{2}+1}$ which is equal to $\frac {(1+n^{-4})^{1/2}} {1+n^{-2}}$. Now take the limit of the numerator and the denominator.

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The limit comparison test is naturally stated in terms of asymptotic equivalence: if $f(n) \sim g(n)$ as $n \to \infty$ then $\sum f(n)$ and $\sum g(n)$ converge or diverge together. The notation $f(n) \sim g(n)$ means $$\lim_{n\to\infty} \frac{f(n)}{g(n)} = 1.$$ In your case, $\sqrt{n^4 + 1} \sim \sqrt{n^4} = n^2$ and $n^3 + n \sim n^3$, so $$\frac{\sqrt{n^4+1}}{n^3 + n} \sim \frac{n^2}{n^3} = \frac{1}{n}.$$ Since the harmonic series diverges, your series does too.

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