3
$\begingroup$

This was supposedly an easy limit, and it is suspiciously similar to a Riemann sum, but I can't quite figure out for what function.

$$\lim_{n\to\infty}{\frac{1}{n} {\sum_{k=3}^{n}{\frac{3}{k^2-k-2}}}}$$

Well, even the fact that $\frac{3}{k^2-k-2} = \frac{1}{k-1}-\frac{1}{k+2}$ doesn't seem to simplify the problem. I thought this would be a telescoping sum, but it's clearly not.

Is that a Riemann sum at all?

$\endgroup$
  • $\begingroup$ Set $k=3,4,5,6$ etc. and add $\endgroup$ – lab bhattacharjee Feb 16 at 14:37
  • $\begingroup$ Your partial fraction decomposition does indeed simplify the problem. Lots of terms are cancelled $\endgroup$ – Gabriel Feb 16 at 14:38
  • $\begingroup$ That is not a Riemann sum. Since the partition width is just $\frac 1n$, a similar Riemann sum would have the form $$\frac 1n \sum f\left(\frac kn\right)$$ for some function $f$. But note that other than the width multiple, the summand depends only on the ratio of $k$ and $n$. This is not the case in your sum. $\endgroup$ – Paul Sinclair Feb 16 at 20:44
  • 2
    $\begingroup$ This is an easy limit--it's zero by inspection. Since the sum is clearly convergent , the $1/n$ term dominates the asymptotic behavior. $\endgroup$ – eyeballfrog Feb 16 at 22:11
8
$\begingroup$

$$\sum_{k=3}^n \frac{3}{k^2-k-2} = \sum_{k=3}^n \frac{1}{k-2}- \sum_{k=3}^n \frac{1}{k+1}$$

You (and I) were mistaken before, see @Romeo 's answer.

Notice that $$\sum_{k=3}^n \frac{1}{k-2}=\sum_{k=0}^{n-3} \frac{1}{k+1}$$

Insert above you get $$\sum_{k=3}^n \frac{3}{k^2-k-2} = \sum_{k=0}^{n-3} \frac{1}{k+1} - \sum_{k=3}^n \frac{1}{k+1} = \sum_{k=0}^2 \frac{1}{k+1} - \sum_{k=n-2}^{n} \frac{1}{k+1}= $$$$=1+\frac{1}{2}+\frac{1}{3} - \frac{1}{n-1}-\frac{1}{n}-\frac{1}{n+1}$$

Of course this argument requires $n\geq 3$.

$\endgroup$
5
$\begingroup$

Isn't it $k^2 -k -2 = (k+1)(k-2)$? In that way, $$ \frac{3}{k^2-k-2} = \frac{1}{k-2} - \frac{1}{k+1} $$ and this is likely to be telescopic.

$\endgroup$
  • $\begingroup$ It looks like the OP already came up with the equality. $\endgroup$ – Yanko Feb 16 at 14:39
  • 2
    $\begingroup$ @Yanko It seems to me the signs of the decomposition of the OP's (and yours) are wrong. Am I mistaken? $\endgroup$ – Romeo Feb 16 at 14:40
  • $\begingroup$ Nope you are correct! $\endgroup$ – Yanko Feb 16 at 14:41
  • 1
    $\begingroup$ @Yanko Thanks a lot, now your post is correct :-) $\endgroup$ – Romeo Feb 16 at 14:42

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.