3
$\begingroup$

To evaluate the limit $\lim \limits_{n \to \infty} \sum_{k=1}^{\lfloor n/2 \rfloor} \frac {n^2}{n^2-k^2}$ I want to use Riemann sums and integrate but I'm having a bit of a hard time as I'm not sure what partition to use. I figured that the relevant function would be $f(x) = \frac{1}{1-x^2}$ but I can't see how to get the partition factor in the mix.

$\endgroup$
  • $\begingroup$ Sum of $[n/2]$ terms each of which is at least $1$? Something is really missing. $\endgroup$ – metamorphy Jun 22 '18 at 9:29
  • $\begingroup$ All terms are smaller than 1, not sure what you mean here. $\endgroup$ – Noa Jun 22 '18 at 9:30
  • $\begingroup$ No, even greater than $1$ actually. A multiplier of $1/n$ would indeed link the question to the matter of integration. $\endgroup$ – metamorphy Jun 22 '18 at 9:35
  • $\begingroup$ Oh, sorry. Silly mistake, got things mixed up. You're correct, I got hung up on this particular idea and forgot to take that into consideration $\endgroup$ – Noa Jun 22 '18 at 9:36
2
$\begingroup$

Actually any term of the sum is $> 1$ and the sequence goes to infinity which is a bit weird here if you are thinking to a Riemann sum. Are you sure that the numerator is $n^2$? Maybe it is $n$.

Note that for $k=1,\dots, \lfloor n/2 \rfloor$, we have that $$0<\frac{1}{n}<\frac{2}{n}<\dots <\frac{k}{n} <\dots<\frac{\lfloor n/2 \rfloor}{n}\leq \frac{1}{2}.$$ Moreover, if the numerator is $n$ then, as $n$ goes to infinity, $$ \sum_{k=1}^{\lfloor n/2 \rfloor} \frac {n}{n^2-k^2}=\left(\frac{1}{n}\sum_{k=1}^{\lfloor n/2 \rfloor} f(k/n)\right)\to \int_0^{1/2}f(x)dx$$ with $f(x) = \frac{1}{1-x^2}$. Can you take it from here?

$\endgroup$
  • $\begingroup$ I get that last equality but I'm not sure what to do about this other $n$ factor. We integrate over the interval $[0,1/2]$ and get a number. Is the limit supposed to be infinite or am I missing something again? $\endgroup$ – Noa Jun 22 '18 at 9:36
  • $\begingroup$ Since any term of the sum is $>1$ the limit is $+\infty$ (without the Riemann sum). If you are thinking to a Riemann sum, maybe the numerator is $n$ and not $n^2$. $\endgroup$ – Robert Z Jun 22 '18 at 9:38
  • $\begingroup$ Okay, I get it now, thanks for your help! $\endgroup$ – Noa Jun 22 '18 at 9:40
2
$\begingroup$

Since $$ \begin{align} \sum_{k=1}^{\lfloor n/2\rfloor}\frac{n^2}{n^2-k^2} &=\sum_{k=1}^{\lfloor n/2\rfloor}\frac1{1-\frac{k^2}{n^2}}\\ &\ge\lfloor n/2\rfloor \end{align} $$ Thus, the limit is infinity.


However, if the question had been $$ \begin{align} \lim_{n\to\infty}\sum_{k=1}^{\lfloor n/2\rfloor}\frac{n}{n^2-k^2} &=\lim_{n\to\infty}\sum_{k=1}^{\lfloor n/2\rfloor}\frac1{1-\frac{k^2}{n^2}}\frac1n\\ &=\int_0^{1/2}\frac{\mathrm{d}x}{1-x^2}\\ &=\frac12\int_0^{1/2}\left(\frac1{1-x}+\frac1{1+x}\right)\mathrm{d}x\\ &=\frac12\left[\log\left(\frac{1+x}{1-x}\right)\right]_0^{1/2}\\[6pt] &=\frac12\log(3) \end{align} $$

$\endgroup$
  • $\begingroup$ Now that I think of it, I may have copied this exercise incorrectly when I was in class and what you're suggesting might have been the correct answer in the first place. $\endgroup$ – Noa Jun 22 '18 at 9:49

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.