# Does this series converge or diverge. Correct use of limit comparison?

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?

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

Hint: $$\dfrac{\sqrt{n^4+1}}{n^3+n} > \dfrac{1}{2n}$$
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.
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.