Convergence of the series $\sum_{i=1}^\infty \sqrt{2n+1}/n^2$ How does the series $\sum_{i=1}^\infty    \sqrt{2n+1}/n^2$  converge? I have yet to receive a result that is not inconclusive. If you could tell me what test you used to confirm its convergence that would be greatly appreciated.
 A: $$\frac{\sqrt{2n+1}}{n^{2}}=\frac{n^{\frac{1}{2}}}{n^{2}}\sqrt{2+\frac{1}{n}}\le\sqrt{3}\frac{1}{n^{\frac{3}{2}}}$$
A: For this problem, you can use the Integral Test for Convergence. Let's first verify the test can be used. To use the test $a_n$ must always be positive, continuous, and decreasing. 
Step 1:
$$\sum_{n=1}^\infty \frac{\sqrt{2n+1}}{n^2} \text{ and } a_n = f(n) \text{ and } f(x) = \frac{\sqrt{2x+1}}{x^2} \tag{initial declarations}$$
Step 2:
$$ f(x) > 0 \Leftarrow\Rightarrow x > 0 \tag{positive for all domain} $$
Step 3:
$$ f(x) \text{ is continuous for all } x > 0 \tag{continuous for all domain} $$
Step 4:
$$ f'(x) = \frac{1}{x^2\sqrt{2x+1}}-\frac{2\sqrt{2x+1}}{x^3} < 0 \Leftarrow\Rightarrow x > 0  \tag{decreasing} $$
Step 5 (Integral Test):
$$ \int_1^\infty \frac{\sqrt{2n+1}}{n^2} dn = \lim_{b\rightarrow\infty}\int_1^b \frac{\sqrt{2n+1}}{n^2} dn = \sqrt{3} - \ln(-(\sqrt{3}-2)) \approx 3.049009$$
Since $\sum_{n=1}^\infty a_n$ meets the conditions for the Integral Test, and the integral of $a_n$ converges, we can deduce that the series also converges.

Evaluation of the integral can also be expressed as:
$$ \int_1^\infty \frac{\sqrt{2n+1}}{n^2}dn = \sqrt{3} + 2\sinh^{-1}(\frac{1}{\sqrt{2}}) \approx 3.049009 $$
See WolframAlpha.
A: $$\dfrac{\sqrt{2n+1}}{n^2}\sim\sqrt{2}\dfrac{\sqrt{n}}{n^2}=\sqrt{2}\dfrac{1}{n\sqrt{n}}.$$
$\sum\dfrac{1}{n^p}$ converges if $p>1$
