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I have a problem where I'm supposed to find if the series converges. I'm looking at the solutions manual and I'm really confused. I know it uses the limit comparison test, but one moment it goes from $n+n^2+n^3$ to $1/n^2+1/n+1$ and I feel like I don't understand how they did that arithmetic.

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They divided both the numerator and the denominator by $n^3$. –  njguliyev Oct 13 '13 at 20:28
    
Ah, I can see how they did that for the numerator, but I don't see it in the denominator. Also, is there any particular reason as to why that's what they would have done? –  FrostyStraw Oct 13 '13 at 20:31

1 Answer 1

$$\large\frac{n+n^2+n^3}{\sqrt{1+n^2+n^6}}\cdot\frac{1/n^3}{1/n^3}=\frac{\frac1{n^2}+\frac1n+1}{\sqrt{\frac{1+n^2+n^6}{n^6}}}=\frac{\frac1{n^2}+\frac1n+1}{\sqrt{\frac1{n^6}+\frac1{n^4}+1}}$$

Remember that $a^3\sqrt{b}=\sqrt{a^6b}$. The point is to reduce both numerator and denominator to expressions with finite limits: those fractions with powers of $n$ in the denominator all tend to $0$ as $n\to\infty$, so the limit can be evaluated by eye.

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