# How can I learn how to solve hard problems like this Example?

Preparing to the Math Olympiad , I found a book and decided to go through it as a summer project. I'm good at math but far from being an Olympian, yet I want to broaden my horizons - problems in this book are not theoretical rather technical and I want to get some new tools.

And guess what? These problems are hard for me. For example:

Solve $$\sum_{n=1}^{2019} \frac{1}{\sqrt{n+\sqrt{n^2-1}}}.$$

I've tried to solve that and gave up after about half an hour. I looked at the answer at the end of the chapter and there was a

HINT:

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

My question is: How does one come up with that second equality? I've checked that it holds and I understand it. But what action (generally speaking) should I take to get from the middle part of the equality to the last on my own?
I'm sure that there is a way to consistently solve problems like this one. (Of course sometimes you need "a spark of genius" to solve a problem, but one question doesn't seem to fall into that category.)

I'm aware that I lack knowledge and experience to get to the solution on my own. How could I get those? Some problems in my book are fairly easy, I would say almost too easy, I solve them in few minutes. On the other hand, others are like problem I mentioned — even if I know what I should be doing, I don't know how.

Should I spend more time on hard ones before giving up? Should I memorize solutions by heart to get techniques? What could I do if I'm stuck? How can I look for techniques to solve problems?

• I'm sure 90% of the people who solved this problem on the actual exam sat for at least 30 minutes just staring at the problem and trying a bunch of different things before reaching the hinted solution. Most people, even experienced Olympians, do not see these solutions immediately. This part of the process is an essential aspect of problem solving that books rarely mention at all. Commented Jul 11 at 11:37
• Given that the hint leads to a telescoping sum (or rather, two "interwoven" telesciping sums), it might be a good idea to collect some intel and try to find the first half a dozen terms in a form as much simplified as possible. Once you have those, the recurrence relation between partial sums may hopefully just stare into your face. -- Of yourse, in the contest situatsion, you do not know the hint, but you can be sure that the result will be somewhat "simple-ish". Thus trying to compute the first few terms is certainly something that may prove useful to get an idea Commented Jul 11 at 12:02
• For math olympiads, telescoping sums are a common thing to try.
– qwr
Commented Jul 12 at 4:04
• There's a "cheat" to this particular question, and it's to recognize how artificial it seems. $2019$ is clearly an arbitrary number and the form of the summand also gives a very suspicious smell. Since you know this is intended to be solvable, you can then deduce that the problem author has a very specific and common trick in math competitions in mind. After some practice, you will remember these tricks and see that telescoping sum is very likely. While it is not obvious how you should tackle summations in general, this summation is almost advertising how it should be solved. Commented Jul 12 at 11:28
• I really, really, really recommend the book The Art and Craft of Problem Solving, by Paul Zeitz (easy to find in full online, but also worth getting hold of a real copy if you can afford it). It’s exactly aimed at teaching the kinds of techniques you’re asking for here (mainly by going through lots of problems, but with more thoughtful explanations of the solutions than usual), and is very well-written and fun to work through. Some of the techniques it gives are rather specific to olympiad-style problems, but others are applicable much more widely as well. Commented Jul 12 at 14:57

Technique : rationalization

When we are given $$A/B$$ or $$1/B$$ , we try to eliminate the Denominator. That will involve "conjugates" , that is , When Denominator is $$x+\sqrt{y}$$ , we could multiply Denominator [ & Numerator ] by $$x-\sqrt{y}$$ to get $$x^2-y$$.
In Contests , that may evaluate to to $$1$$ , like here. Denominator gone !

[[ I will add the calculation for your case here , in case you are not able to see it. ]]

Technique : root of Surd is a Surd

In general terms : When we are given $$\sqrt{A+\sqrt{B}}$$ we try to make it $$C+\sqrt{D}$$
In other words , $$[C+\sqrt{D}]^2=A+\sqrt{B}$$
That gives us $$C^2+D+2C\sqrt{D}=A+\sqrt{B}$$
Hence we get $$C^2+D=A$$ , $$2C\sqrt{D}=\sqrt{B}$$

[[ I will add the calculation for your case here , in case you are not able to see it. ]]

Technique : Prepare Prepare Prepare

When you see a Solution , try to figure out what general technique was used & see how you can apply it elsewhere.
Make your own Exercise employing that technique.
Make variations.
When you come across a new Exercise , check which known technique can be used there.
If you are not aware , no worries , you are learning. Check the Solution , add that new technique to your tool set.
Soon you will have a large tool set.

Prepare Prepare Prepare

Technique : Summation via telescoping

Suppose we have $$S=\Sigma [X_n]$$ & we are somehow able to make it $$S=\Sigma [Y_{n+1}-Y_n]$$ , then write it out like this :
$$S=[Y_1-Y_0]+[Y_2-Y_1]+[Y_3-Y_2]+\cdots+[Y_{2019}-Y_{2018}]$$
We can see that all terms in the middle will cancel , leaving Exactly 2 terms.
Hence $$S=Y_{2019}-Y_0$$

[[ I will add the calculation for your case here , in case you are not able to see it. ]]

This Summation technique is known as telescoping.
Make variations to play with.
Experience will come with time & effort.
Practice is Key !

And guess what, these problems are hard for me.

• Should I spend more time on hard ones before giving up?
• Should I memorise solutions by heart to get techniques?
• What could I do if I get stuck?

I am afraid there is no easy answer to questions like this. Hard work and innate ability are both ingredients, but nobody will be able to tell you how hard you have to work to solve a problem, or how difficult a problem you will be able to solve on your own (and after all, there is a limit, because you only live so long).

Memorisation can help, but many (most?) successful problem solvers do not memorise very much. Practice is ten times as important as memorisation. And figuring out how to solve a problem on your own is ten times as valuable as looking up a solution. But sometimes it will take you more than ten times as long to discover the solution on your own, and you are only human. We all look up a solution from time to time. But you should try as hard as you can to make it an exception rather than a rule.

There are some general rules, but trying to be too precise misses the point. There is no general algorithm for problem solving. People mostly just learn by doing. You find tricks you like, and you try applying them to new problems you find. Sometimes the trick is good and you keep it around. Richard Feynman famously got a reputation for doing hard integrals, when on many occasions he was just applying a trick nobody else knew about. Some tricks are so well-known to be useful we put them in books and teach classes about them.

I'll be honest with you and say that the only thing that jumps out at me when I look at that fraction is "difference of two squares". The thinking part of my brain wants to say something like "I have no idea how to add up a bunch of square roots, so I had better find some way to express $$n-\sqrt{n^2-1}$$ as a square of something".

Difference of two squares tells me $$n^2-1 = (n+1)(n-1)$$. Maybe that's useful maybe it isn't. It's not a square, which sucks. What would a square look like? Well $$(\sqrt{n}+X)^2 = n+2\sqrt{n}X + X^2$$. I picked $$\sqrt{n}$$ so I'd get an $$n$$ term there. Another thing I noticed is that $$(n+1) + (n-1) = 2n$$. In fact that $$\sqrt{n}$$ is reminding me of something I've seen before.

If you add two square roots and square it, the "square-root-ness" goes down. What do I mean? This identity:

$$(\sqrt{a}+\sqrt{b})^2 = a+b + 2\sqrt{ab}$$

You use this sometimes if you want to solve an equation like $$\sqrt{x+1} + \sqrt{x} = 1$$. Anyway in this case we can pattern match a bit and say "I want to make $$ab$$ like $$n^2-1$$". Well what about $$a=n-1$$ and $$b=n+1$$. Well then $$a+b=2n$$. Oh, we're done:

$$n+\sqrt{n^2-1} = (\sqrt{n-1}+\sqrt{n+1})^2/2$$

Wait! I misread the question! There's actually a minus sign, but... the minus sign doesn't make a difference. We're done! The sum telescopes so you can work out the rest easily.

There isn't a magic formula that will tell you how to do this. You just figure it out by practicing a lot. Most people can do this sort of problem if they work at it. You probably have a limit, but it's also probably higher than you think it is. There are words like "generalisation" and "abstraction" that people could use to document what was going through my head when I solved this problem. But it's not really how you learn to solve the problem. You learn by doing, and by association.

## Addendum: There really is no algorithm

I noticed this while proof reading my answer, and I couldn't resist pointing it out. I gave an example earlier of an equation $$\sqrt{x+1} + \sqrt{x} = 1$$. I promise I picked it randomly. People sometimes try to teach you that there are simplifications you should always do. Rationalising the denominator is one (you used it in your question). But this is a good example where you should do the exact opposite! By "un-rationalisting the denominator", you get:

$$\frac{1}{\sqrt{x+1} - \sqrt{x}} = 1 \implies \sqrt{x+1}-\sqrt{x} = 1$$

and not much more work is needed to get a solution. Much easier than repeated squaring!

If you gave up after half an hour, you didn't really try.

Prem and others have already posted some concrete tips on how this particular problem could be approached, so I'll address the more general question.

First step is to look at the problem and see if it looks familiar, if you have solved something similar before, or some idea on how to solve it pops into mind. You then try those things out, and see if it either solves it or turns it into a simpler problem. That is the quick and easy approach, but mostly only works on simple problems that are already familiar.

Some times, a problem can be solved using only one idea or method, but that's the typical textbook kind of problem where the textbook also teaches you how to solve them. Other times, you need to combine several steps, but where each step makes the problem simpler.

If you have solved a lot of problems, you will have seen a lot of tricks and techniques, and more easily have several ideas on things to try. However, often you may need variations on those techniques, not just replicate a previously used technique. Practice can also teach you when a problem has been made simpler.

This problem needed two steps, where the first is pretty standard to get rid of the denominator, while the second was more of a variation on familiar themes. You already have a major hint in the problem text, that there is a simple solution, which means there must be some simplification available. Eg:

• Try to simplify square roots like what Prem suggested, but you may have to try more than one approach.
• Can each term we simplified to the form $$a_n-b_n$$ in such a way that the $$b$$s get canceled against $$a$$s?
• Compute some terms, worst case you do it numerically, and see if you can guess at what the answer may be. And by guess, I mean, try lots of things to see if it matches.

This is how you solve relatively simple problems and problems that can be solved using existing methods. That does not mean that there is a textbook solution or a textbook approach to solving it, just that with enough practice and work, the solution can be found.

Some times, there are multiple steps in the solution, where some of the steps do not immediately seem to result in simplifying the problem. These are much harder to solve and sometimes require some degree of inspiration on top of hard work, but with practice you can also learn to recognise that a harder version of the problem may lend itself to other techniques and approaches.

If you run out of ideas, but know there is supposed to exist a relatively simple solution, you put it aside and revisit it later.

While reading solutions may be helpful in learning tricks an techniques, the only way to learn when and how to apply them is to actually work on solving problems yourself. The risk of going for the solution too early is that the next technique you need may be more difficult to apply, and so you and up learning lots of tricks, but do not learn to recognise when and how to apply them.

• Sometimes , "some tough love" is very much necessary ! Hence +1 for that & the other tips too.
– Prem
Commented Jul 11 at 13:01
• As a concrete example of where a step makes the problem more difficult, but suddenly opens up a new technique that you couldn't to before, consider the statement "for any $n$, we have $\sum_{k=1}^n\frac1{n^2}<2$". It's not immediately obvious how one would attack this, but the moment we pose the strictly stronger result with $\leq 2-\frac1n$ in place of $<2$, induction works directly. Commented Jul 11 at 13:21
• In grade school they basically taught us not to think for extended periods of time, because all the questions are formulaic. I like to believe if I had less useless classes, I could've spent more time on extended math thinking...
– qwr
Commented Jul 12 at 4:06
• > "If you gave up after half an hour, you didn't really try." I dunno that that's true. In a competition, sure, you should work problems until the time is up. But as practice, if you're working stuff for 30 minutes and getting somewhere you should keep working, but if you're staring at it for 30 minutes and don't even know where to start, it's a sign to either 1. get some sleep because your brain isn't working, 2. find an easier problem on a similar theme, or 3. do some research, because you don't have the tools you need yet.
– Kaia
Commented Jul 12 at 18:46

I have a few suggestions for you

1. Spend more time on each problem. In math olympiads, you should NOT be just giving up after 30 minutes. Math olympiad problems are designed so that an they take a few hours to figure out and solve. Keep thinking. When using this time, try many different methods.

2. Do more problems. The more problems you do, the better you will get, and the easier you will recognize what you should do given a certain problem.

3. Do easier problems than these. You say "I'm good at math but far from being an Olympian". Doing these type of problems at this level will only discourage you. This particular problem is quite simple for an Olympian. Since you can not solve it in 30 minutes, try doing simpler questions that are not Olympiad level. For example, the AMC and AIME series of questions are excellent. Simpler questions allow you to learn methods of solving problems in less time than doing Math Olympiad problems.

For the particular problem you mentioned, I can give you a few things that you can use to solve the problem.

1. 2019 is a really random number. In most (but certainly not all) cases, whenever a problem includes a year, it is most likely random. In those cases, try ignoring that number, and focus on solving the general case,

2. Since the square root is in the denominater, simply rationalize the denominator, to get the first equality.

3. After the first equality, you a nested square root(A square root inside a square root). There is a very standard method to solving it.

Overall, since you did not notice this solution, to me, that indicates a lack of experience. All steps in the solution are very natural to an experienced problem solver.