# convergence using comparison test

I am trying to show that the series $\displaystyle\sum_{n=1}^{\infty} \dfrac{\sqrt{2n+1}}{n^{2}}$ is convergent. Mathematica says by the comparison test the series is convergent but it doesn't say what comparison it makes.

I know the series $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^{2}}$ is convergent but the original series is greater than this series for every $n$ so it doesn't work for the comparison.

I need to find a converging series that is greater than the original series to compare to. How do I find the series for comparison?

## 2 Answers

$$\sum_{n=1}^{\infty} \dfrac{\sqrt{2n+1}}{n^{2}} \leq \sum_{n=1}^{\infty} \dfrac{\sqrt{3n}}{n^{2}}$$ and now convergence should be obvious.

• Thanks! It does seem obvious now – Devin Crossman Nov 18 '14 at 3:32

$\sqrt{2n+1}\leq\sqrt{3n}$ for all integer $n\geq 1$, then $$\frac{\sqrt{2n+1}}{n^2}\leq\frac{\sqrt{3}}{n^{3/2}}$$ By comparison test with the $p$ series, we have $\displaystyle{\sum_1^{\infty}\frac{\sqrt{2n+1}}{n^2}}$ converges.