# Comparison test

Use the comparison test to determine whether or not

$$\sum\limits_{n=1}^\infty \dfrac{\sqrt{n}}{1+n+n^2}$$

converges.

Would I use $\frac{1}{n^\frac{3}{2}}$ for the test and then show,

$$\lim_{x \to \infty} \dfrac{\sqrt{n}}{1+n+n^2}\frac{n^\frac{3}{2}}{1}=\lim_{x \to \infty} \dfrac{n^2}{1+n+n^2} = 1$$

$$\sum\limits_{n=1}^\infty \frac{1}{n^\frac{3}{2}}\Rightarrow \sum\limits_{n=1}^\infty \dfrac{\sqrt{n}}{1+n+n^2}\Rightarrow converges$$

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I think you applied the limit test, not the comparison test. – DonAntonio Dec 5 '12 at 4:18
@DonAntonio: nope, that's comparison the way it's taught in many places to play around the students' lack of inequality ability. If $a_n/b_n$ has finite limit, then you have comparison between their respective series. – Martin Argerami Dec 5 '12 at 4:34
Perhaps so, @MartinArgerami, yet I think in most places the comparison test is what I did in my answer below. Some places, like Wiki, is what show you, and ditinguish between this comprison test, calling it direct comparison test, andthe limit comparison test. – DonAntonio Dec 5 '12 at 4:43
I learned the same distinctions that DonAntonio did between the two tests. – Joe Dec 5 '12 at 4:44

I want to clear up something and sort of add a note on DonAntonio's answer. It seems you're saying your original sum is equal to $\sum_{n=0}^{\infty}\dfrac{1}{n^{3/2}}$. Instead, it really is similar to, but not equal to. Thus, we should use the Comparison Test and watch our inequalities.

Looking at behavior near $\infty$ (that is, looking at the terms that dominate), we have that: $$\frac{\sqrt{n}}{1+n+n^2} \thicksim \frac{\sqrt{n}}{n^2} = \dfrac{1}{n^{3/2}}$$

Note that $$\frac{\sqrt{n}}{1+n+n^2} \le \dfrac{1}{n^{3/2}} \forall n \in \mathbb N$$

Thus we can say our original sum is less than $\sum_{n=0}^{\infty}\dfrac{1}{n^{3/2}}$ which is a convergent p-series ($p>1$) which means our original sum converges by comparison.

It appears you used the Limit Comparison Test. For the LCT, we have an assumption that $$\lim_{n\to\infty} \frac{a_n}{b_n}$$ exists and equals $L$. Then we have two cases: $$L>0 \implies \sum{a_n} \ \ \ \text{converges} \iff \sum{b_n} \ \ \ \text{converges}$$ and for the case $L=0$ : if $\sum{b_n}$ converges, then $\sum{a_n}$ converges.

Note that the second case only works one way. Your reasoning using the LCT seems fine though by stating $\sum{a_n}$ converges because $L>0$ and $\sum{b_n}$ is a convergent p-series.

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Thanks for the detailed explanation, I now understand the difference between the two tests. – user51462 Dec 5 '12 at 4:47
No problem, glad to help! – Joe Dec 5 '12 at 4:48

$$\frac{\sqrt n}{1+n+n^2}\leq \frac{\sqrt n}{n^2}=\frac{1}{n^{3/2}}$$

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Thanks for clearing that up – user51462 Dec 5 '12 at 4:27

Your answer is correct if you can assume the knowledge that $\sum_n n^{-3/2}<\infty$.

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