Michael K. answered • 03/08/20

Former Mathlete turned Startup Dev and Tutor

Important background knowledge:

sin(x) is about equal to x when x is small (and the approximation gets better as x nears zero).

n^-2 = 1 / n^2 gets very small as n gets large (it converges to zero; 1 / infinity = 0)

Raising a tiny number to the third power results in an even tinier number (any time the number is smaller than 1, cubing it will get you a number closer to zero; e.g. half of a half of a half is smaller than a half)

This sequence is the result of taking the sequence an = n^-2, then

- multiplying each term by (-1)^n
- taking the sine of each term of the result of that
- cubing each term of the result of that

So we have:

0. The sequence n^-2 gets smaller and smaller and converges to zero.

- Multiplying each term of a sequence that converges to zero by 1 or negative 1 (which is what happens when we multiply by (-1)^n) will give a sequence that also converges to zero. We can flip what side of zero any given term is on, but we are still getting closer and closer to zero (and therefore converging to zero) regardless of what sign we have.
- Taking the sine of each term of a sequence that converges to zero will give a sequence that also converges to zero, because when the quantity is very small (positive or negative), taking the sine leaves it essentially unchanged.
- Finally, cubing each term of a sequence that converges to zero will result in a sequence that converges to zero because the cube of a number close to zero will be even closer to zero.

So the final sequence converges to zero, as was to be shown. Fun fact: QED just means "as was to be shown"; if you see it at the end of math proofs it just means we have now proven what we set out to prove.

Additional explanations / clarifications:

If we're measuring in degrees instead of radians, sin(x) will be close to x * a constant factor instead of x * 1, but multiplying a sequence converging to zero by a constant still gives a sequence that converges to zero, so it turns out not to matter.

It may help to realize that because sin(x) and x^3 are odd functions (which just means that for any x, sin(-x) = -sin(x) and (-x)^3 = -(x^3)),

sin^3((-1)^n * n^-2) = (-1)^n * sin^3(n^-2)

So if the multiplication by (-1)^n step is confusing with how it affects the sine function, you can isolate it and leave it for the last step.

A more rigorous version of multiplying by (-1)^n not mattering:

- If you always multiplied by 1, the sequence would still converge to zero. (it would be a smaller and smaller positive number).
- If you always multiplied by -1, the sequence would still converge to zero. (it would be a tinier and tinier negative number).
- The actual sequence we got never has a value greater than the positive sequence or less than the negative sequence. We say it is bounded above by the positive sequence and below by the negative.
- Since our sequence is bounded above and below by two sequences that individually converge to zero, our sequence also converges to zero (picture boundaries giving a tighter and tighter window around zero that our function must stay within).