An unusual way of telescoping

Middle school was a flower in spring:  It was beautiful, it bloomed, yet it wilts momentarily and leaves behind the barren, untamed, yet still hopeful landscape.  It was a lot, yet it was too little that I had done in that time.  One of those middle school things that I continued to cherish was those middle school math contests.

Scroll to the "The Problem" sections for the problem I'm writing up.  "The Real Solution" sections is no-fluff for those who don't care about my stupidity and just want to see some advanced telescoping.

One of the largest of those (if not the largest) math contests for middle schoolers was MATHCOUNTS; participants were branded as Mathletes (pun on Math and Athletes).  I've blogged a bit about the end of my career as a Mathlete in my 8 tetracubes article.  It definitely wasn't one of those and-they-all-lived-happily-ever-after fairy-tale endings, but it was... fine (in the usual, not slang sense of the term).  But still, I flunked Sprint #30 and failed spectacularly, realizing what went wrong only 20 seconds after they call time (and lost my bragging rights).  Sprint #30 was pretty much the end of my career (as the target round is just a special snowflake [0]), and I spent the last 6 or so minutes of the test or so grinding it out (bad time management...); nearly solving it left an empty gap deep in my heart.

And below will be my attempt to recover that bit of triumph from having solved the problem, to ease that bit of sadness that will continue to persist into my high school years.

The problem

Let an integer $n>0$ be chosen with probability$$\Pr(n)=\frac{1}{n(n+1)}.$$What is the probability that $n$ is $1$ less than a triangular number?

Freely paraphrased from MATHCOUNTS 2026 State, Sprint #30

Yes, this was the problem.  That simple.  So... let's... Just Do It?

The $m$-th triangular number is $\frac{m(m+1)}2$, so $1$ less than the $m$-th triangular number is $\frac{m(m+1)}2-1$.

Now, note that

$$n=\frac{m(m+1)}2-1=\frac{m(m+1)-2}2=\frac{m^2+m-2}{2}=\frac{(m+2)(m-1)}2.$$

Although this just looks like a simple factoring, in contest I somehow did not have the idea to expand $m(m+1)$ before refactoring into $(m+2)(m-1)$ for later use in partial fraction decomposition, and was stuck staring at how to handle $m(m+1)-2$ for 1 minute (or a bit more).  Oh well.  We also know that $n+1$ cancels out the $-1$ part in $n$ and is just $\frac{m(m+1)}2$.

Now I know that the original $\Pr(n)$ can be decomposed into 2 fractions for telescoping (as is usual for $\frac1{n(n+1)}$.  Great, now I realized.  We have the following if we do that and then plug in $n$ and $n+1$ we got previously:

$$\Pr(n)=\frac{1}{n(n+1)}=\frac1n-\frac1{n+1}=\frac1{\frac{(m+2)(m-1)}2} - \frac1{\frac{m(m+1)}2}=2\!\left(\frac1{(m+2)(m-1)} - \frac1{m(m+1)}\right)$$

Alright, doesn't look too bad, we can decompose $\frac1{(m+2)(m-1)}$ and $\frac1{m(m+1)}$ once again.  Except, well, if you wanna decompose $\frac1{(m+2)(m-1)}$... you might mess up.  If you write $\frac1{m-1}-\frac1{m+2}$, you'll end up with a numerator of $3$ so it's $\mathbf{\frac13}(\frac1{m-1}-\frac1{m+2})$.  Yes, I messed this up in contest again, but thankfully (or not) wasted only $1$ minute.  Decomposing $\frac1{m(m+1)}$ is trivial, so

$$2\!\left(\frac1{(m+2)(m-1)} - \frac1{m(m+1)}\right)=2\!\left(\frac13\left(\frac{1}{m-1}-\frac1{m+2}\right)-\left(\frac1m-\frac1{m+1}\right)\right).$$

Now, let's start summing this stuff with $m$ as our index -- we need to cover the sum of all probabilities that is $1$ less than the $m$-th triangular number -- we simplify the final expression just a bit further from the above:

$$\sum_{m=1}^\infty \frac23\left(\frac{1}{m-1}-\frac1{m+2}\right)-2\left(\frac1m-\frac1{m+1}\right)=\frac23\sum_{m=1}^\infty \left(\frac{1}{m-1}-\frac1{m+2}\right)-2\sum_{m=1}^\infty\left(\frac1m-\frac1{m+1}\right).$$

And surely, any sane person will notice an error here:  we must have $m\geq2$.  This is because $1$ less than the $m=1$-st triangular number, $1$, is $0$ which is not a positive integer and thus cannot be picked.  Uno!  (Now, draw 25 cards or rework this whole problem.)  3 minutes wasted in contest finding out whatever the heck I got wrong here.

Let's rework.

$$\begin{aligned}&\mathrel{\phantom=}\sum_{m=\mathbf 2}^\infty \frac23\left(\frac{1}{m-1}-\frac1{m+2}\right)-2\left(\frac1m-\frac1{m+1}\right)\\&=\frac23\sum_{m=\mathbf 2}^\infty \left(\frac{1}{m-1}-\frac1{m+2}\right)-2\sum_{m=\mathbf 2}^\infty\left(\frac1m-\frac1{m+1}\right).\\&=\frac23\sum_{k=1}^\infty \left(\frac{1}{k}-\frac1{k+3}\right)-2\sum_{m=\mathbf 2}^\infty\left(\frac1m-\frac1{m+1}\right)\qquad\text{let $k=m-1$ on first term}\end{aligned}$$

Oh wait... was the middle a $+$ or a $-$?  I'm not sure how much time I wasted getting all the signs wrong in contest (probably another $3$ or so minutes). [1]

So anyways, if you just write out the first few terms and see things cancel out, you get $\sum_{k=1}^\infty \left(\frac{1}{k}-\frac1{k+3}\right)=\frac{11}6$ and $\sum_{m=\mathbf 2}^\infty\left(\frac1m-\frac1{m+1}\right)=\frac12$.  So $\frac{2}{3}\cdot\frac{11}{6}-2\cdot\frac12=\boxed{\textstyle\frac29}$.

So, you've seen me mess up several times on solving this problem that should've been easy, given I knew what telescoping was.  I knew how to do it... or, did I?

The real solution

Now, all of the above was just my stupidity.

Why do I say that?  Well, when I looked at the solution packet for this problem, I was baffled.

It turns out that instead of looking to immediately decompose $\frac1{n(n+1})$, if I was just a little more patient, and substituted my $m$ representations of $n$ and $n+1$ first, I would get this:

$$\Pr(n)=\frac{1}{n(n+1)}=\frac{1}{\frac{(m-1)(m+2)}2\frac{m(m+1)}2}=\frac4{(m-1)m(m+1)(m+2)}.$$

They say it's possible to telescope this directly.  How the heck?

Well, note that to telescope, you aren't limited to just $\frac1a-\frac1{a+c}$ where $c$ is a constant.  There can be more complicated things in the denominator.  In this case, we can write

$$\begin{aligned} &\mathrel{\phantom{=}}\sum_{m=2}^\infty \frac{4}{(m-1)m(m+1)(m+2)} \\ &=\sum_{m=2}^\infty \frac{1}{m(m+1)}\cdot\frac{4}{(m-1)(m+2)} \\ &=\sum_{m=2}^\infty \frac{1}{m(m+1)}\cdot \frac{4}{3}\cdot \left(\frac{1}{m-1}-\frac{1}{m+2}\right) \\ &=\frac{4}{3}\sum_{m=2}^\infty \frac{1}{m(m+1)}\left(\frac{1}{m-1}-\frac{1}{m+2}\right) \\ &=\frac{4}{3}\sum_{m=2}^\infty\left(\frac{1}{(m-1)m(m+1)}-\frac{1}{m(m+1)(m+2)}\right) \end{aligned}$$

Note how we just need to decompose $2$ things in the denominator, and just leave the others alone; at last, we can put them back in, and $\frac{1}{m(m+1)(m+2)}$ is precisely the $\frac{1}{(m-1)m(m+1)}$ of the next term.  Now we can telescope, and everything except for the first $\frac{1}{m(m+1)(m+2)}$ cancels out, which is $\frac1{1\cdot2\cdot3}$.  Multiplying that by the $\frac43$ we have in front of the sum we get $\frac29$, the same answer as before.

The key here is that the things you need to split should be the first and the last in the multiplied sequence in the denominator, so that the neat pattern of subtracting the added term happens (as we offset exactly by $1$).

What a fool I was to not know that you can telescope with complicated things in the denominator as long as they eventually cancel out.  It takes like 1 minute to do it the above way if you don't mess up and know what you're doing (I messed up again when writing this post, fixed it, and it took 5 minutes to do the math and write it up, and 3.5 of those minutes are like stupid $\LaTeX$.)

So, there we go:  5 minutes wasted on a Sprint test, 5 minutes that I'll never be able to get back, because it was my last Sprint test of my middle school MATHCOUNTS career.........

So what?

When people asked if I knew what telescoping was, I've always responded, "yes".

But I only learnt of the other technique to solve this problem today—I was unblissfully ignorant towards more complicated forms of telescoping.

I could do the superficial thing, but I clearly did not have the understanding to do even a slight variant of it.

So did I really know telescoping?

It seems way more likely that I'm just a bluff and have no idea about what I'm doing in life either.........

Footnotes and References

[0] Yes, it takes forever to pass out and collect each pair of problems!  It takes 30 minutes to administer a 16-minute test...

[1] I actually did this several times before getting it right for this blogpost.  Still haven't learnt, I guess...