# Determine if the series $\sum_{n=1}^\infty {1\over \sqrt{2n-1}}$ is convergent or divergent

I'm trying to solve the problem in the title and I think I have it narrowed down to using a comparison test, specifically the limit comparison test, which is stated as:

What I don't know though is how to find $b_n$. I know that $a_n={1\over \sqrt{2n-1}}$. Would $b_n$ equal ${1\over \sqrt{2n}}$? If so, my next step would be to find the limit of $a_n\over b_n$, correct?

Or should I only use the p-series test?

Thanks for any help.

Yes, and that limit is $\frac{\sqrt{2n}}{\sqrt{2n-1}}=\frac{1}{\sqrt{1-1/2n}}\to1$.

Therefore the series of $\frac{1}{\sqrt{2n+1}}$ diverges, since $\frac{1}{\sqrt{2n}}\geq\frac{1}{n}$, and the series of $\frac{1}{n}$ diverges.

• Where did $1\over n$ come from exactly? I keep getting confused at that. Thanks for your help. Couldn't I also do this using the P-series test? – JustHeavy Oct 28 '17 at 18:43
• $1/n$ is the harmonic series and is often used as a reference point in limit comparison – Lanier Freeman Oct 28 '17 at 18:44
• Thanks, @LanierFreeman, I feel like using the p-series test is best then since I do not know how to use the harmonic series, I have not learned to use to prove this stuff. Doing it this way then, $a_n = {1\over{\sqrt{2n-1}}}$ and $b_n={1\over{\sqrt{2n}}}$. Now, since $b_n={1\over{(2n)^{1/2}}}$ That means that $p<1$ and it diverges. Is that correct? It seems too easy. – JustHeavy Oct 28 '17 at 18:54

You can use whatever you like. You can also prove tight approximations for the partial sums by noticing that for any $n\geq 1$ the ratio $\frac{1}{\sqrt{2n-1}}$ is bounded between $\sqrt{2n+1}-\sqrt{2n-1}$ and $\sqrt{2n}-\sqrt{2n-2}$, hence

$$\sum_{n=1}^{N}\frac{1}{\sqrt{2n-1}}\in\left[\sqrt{2N+1}-1,\sqrt{2N}\right].$$

• @JackDAurizio Thanks for the response, can you check to make sure I did this correctly. I'm going to use the comparison test. Doing it this way then, $a_n = {1\over{\sqrt{2n-1}}}$ and $b_n={1\over{\sqrt{2n}}}$. Now, since $b_n={1\over{(2n)^{1/2}}}$ That means that $p<1$ and it diverges. Is that correct? It seems too easy. – JustHeavy Oct 28 '17 at 19:15
• @DevHeavy: asymptotic comparison and the $p$-test provide a simple way for proving divergence, that is correct. – Jack D'Aurizio Oct 28 '17 at 19:22