# Is this a Riemann sum (if so, I can't figure out which one)?

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?

• Set $k=3,4,5,6$ etc. and add – lab bhattacharjee Feb 16 at 14:37
• Your partial fraction decomposition does indeed simplify the problem. Lots of terms are cancelled – Gabriel Ribeiro Feb 16 at 14:38
• 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. – Paul Sinclair Feb 16 at 20:44
• This is an easy limit--it's zero by inspection. Since the sum is clearly convergent , the $1/n$ term dominates the asymptotic behavior. – eyeballfrog Feb 16 at 22:11

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

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 $$n\geq 3$$.

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.

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