# Does the series converge or diverge?

Let $$a_n=\frac{\sqrt{3n+1}}{n^2}$$ I cannot find a suitable $b_n$ to use for the comparison test, and when I try to use the ratio test it really becomes a mess and I cannot find the limit as a result. I am going to guess there is something simple I am missing.

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Do you want the convergence of the series $\sum_n \dfrac{\sqrt{3n+1}}{n^2}$ or the sequence $\dfrac{\sqrt{3n+1}}{n^2}$? – user17762 Jun 23 '12 at 23:47

I am assuming that you are interested in the convergence of the series $\sum_n a_n$. Either way the same argument will work.

HINT First note that $\forall n \geq 1$, we have $0 \leq 3n+1 \leq 4n$. Now can you bound the terms and argue?

Note that $\forall n \geq 1$, we have $0 \leq 3n+1 \leq 4n$. Hence, we have that $$0 \leq \sum_n \dfrac{\sqrt{3n+1}}{n^2} \leq \sum_n \dfrac{\sqrt{4n}}{n^2} = \sum_n \dfrac2{n^{3/2}}$$ Recall that $\displaystyle \sum_n \dfrac1{n^p}$ converges for all $p > 1$. Hence, we have that $\sum_n a_n$ converges. You may look here for proofs of $\displaystyle \sum_n \dfrac1{n^p}$ converges for all $p > 1$.

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You can use the limit comparison test. First note that $${\sqrt{3n+1}\over n^2} \sim {\sqrt{3} \over n^{3/2}}.$$ The p-series theorem says that the series of terms on the right converges. By limit comparison, your original series converges.

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If the sequence is $a_{n} = \frac{\sqrt{3n+1}}{n^2}$, observe that:

$$0 \leq \frac{\sqrt{3n+1}}{n^2} \leq \frac{\sqrt{4n}}{n^2} = \frac{2}{n^{1.5}}$$

Hence $\lim_{n \rightarrow \infty} a_{n} = 0$.

If the sequence is $a_{n} = \sum_{n}{\frac{\sqrt{3n+1}}{n^2}}$,

$$a_{n} \leq \sum_{n}{\frac{\sqrt{4n}}{n^2}} = \sum_{n}{\frac{2}{n^{1.5}}} := \sum_{n}{b_{n}}$$

The ratio test does not work in this case since:

$$\lim_{n} \frac{b_{n}}{b_{n+1}} = \lim_{n} \frac{n^{1.5}}{(n+1)^{1.5}} = 1$$

An alternative is the integral test. Observe that:

$$\int_{1}^{\infty}{\frac{2}{x^{1.5}}dx} < \infty$$

Hence, $\sum_{n}{\frac{\sqrt{3n+1}}{n^2}} < \infty$.

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