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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.

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Try comparing to $\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ using the Limit Comparison Test. – user84413 Jul 14 '14 at 14:38
Yes, the limit comparison test using this comparison series also works, since the comparison series is convergent, and $\lim_{n\rightarrow\infty} \frac{a_n}{b_n}$ is finite ($\sqrt{2}$). – MrCMedlin Jul 14 '14 at 15:50
up vote 5 down vote accepted


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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.

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How did you evaluate this integral? Notice that $\ln(-\sqrt{3}-2)$ is undefined. – user84413 Jul 14 '14 at 14:35
Sorry, I left out the parentheses. It should be $\ln(-(\sqrt{3}-2))$. I'll fix it. – MrCMedlin Jul 14 '14 at 15:09
Thanks for your reply; it looks better now. – user84413 Jul 14 '14 at 15:25


$\sum\dfrac{1}{n^p}$ converges if $p>1$

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