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Shades of Fermat's Last Theorem

The familiar Pythagorean 3-4-5 triple gives one solution to (x-1)^n + x^n = (x+1)^n so what about other solutions for x an integer and n= 2, 3, 4 or 5?

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Exhaustion

Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2

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Code to Zero

Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135.

Diverging

Stage: 5 Challenge Level: Challenge Level:1

The first part of the question asks you to show that for natural numbers $x$ and $y$ if ${x\over y}> 1$ then

$${x\over y}> {(x+1)\over(y+1)}> 1.$$

Here's a hint for this: try starting with $x> y$, which you are given, and adding $xy$ to both sides of the inequality.

For the next part of the question you are given a product $P$ and the hint to consider $P^2$ and clearly the first part of the question should come in useful. Look out for a 'magic concertina' effect!!

If you can prove the second inequality then you will have shown that $P$ gets bigger and bigger without limit as you put more terms into the product which proves that the product diverges, hence the title of the question!

For the last bit of the question, taking $k=100$ and repeating the last trick leads to the disappointing conclusion that $Q^2> 101$, this estimate is not good enough.

Go back to the drawing board and do some neat estimating of $Q^2$ calculating the product of the first few terms exactly and using the concertina method on the rest. This will quickly give you the result. This is a good illustration of what mathematicians do all the time with inequalities. They go on sharpening them to get better and better estimates until they get close enough for their purposes.