3
$\begingroup$

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:

enter image description here

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – JustHeavy Oct 28 '17 at 18:43
  • $\begingroup$ $1/n$ is the harmonic series and is often used as a reference point in limit comparison $\endgroup$ – Lanier Freeman Oct 28 '17 at 18:44
  • $\begingroup$ 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. $\endgroup$ – JustHeavy Oct 28 '17 at 18:54
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ @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. $\endgroup$ – JustHeavy Oct 28 '17 at 19:15
  • $\begingroup$ @DevHeavy: asymptotic comparison and the $p$-test provide a simple way for proving divergence, that is correct. $\endgroup$ – Jack D'Aurizio Oct 28 '17 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.