PAT Maths question but I think I can do better?

I'll let the picture do the talking


The "markscheme" answer is from at least 2 different sources so :\

So yeah am I right or not?

The RHS does not exist for 0 < x < 1.

"the other rules"? Note that for $0<x\le 1$, $\sum \frac 1 {x^n}$ diverges, as you had previously said.

Why not? It might be undefined because it converges ... but surely to a mathematician that's good enough right?

I basically meant the other stuff I said. If something diverges that means it gets pretty huwowage so surely that means it'd be bigger than the measly x+2 ?

I'm not sure what you're trying to say here.

I guess the point is this:

1. You have summed the series to find that $S=\frac{x}{x-1}$. However that sum only converges to that expression for $x <-1, x>1$. The sum does not spit out the same value as the expression for values of x in the range $-1 <x<1$ so they are not the same thing in that range e.g. $x=0.1 \Rightarrow S=\frac{0.1}{0.1-1}=-\frac{1}{9}$ but the series does not give that value

2. You have done some algebraic inequality solving with the expression for S to find that $-\sqrt{2}<x<\sqrt{2}$. However, we are working firstly only with +ve x so $0 < x$, and that expression is only a valid representation of the sum if the +ve value of x is in the range $x>1$, and hence your inequality solution is only valid for values of x in that range.

So putting those points together we get the markscheme result.

The expression $x <y$ can only be assigned a true or false value if both $x,y$ are real numbers. $\infty$ is not a real number so the expression $x < \infty$ is not a valid expression - it's not well-formed. In order to decide for which values of $x$ the following expression is true:

$E_1=x+2<1+\frac{1}{x}+\frac{1}{x^2}+\cdots=E_2$

we must restrict ourselves to values of +ve $x$ where both $E_1,E_2 \in \mathbb{R}$ else the expression is not well-formed.

If $0<x<1$ the expression $E_2$ is not equal to any real number, so we can't work with values in that range at all. For values of $x>1$ then $E_2$ is correctly represented by $x/(x-1)$, so we replace it with that expression which allows us to solve the inequality easily with no need to play around with an infinite series.

Take x=2.

RHS = 1 + 1/2 + 1/4 + 1/8 + ... = 2

Your formula claims it converges to x/(x+1) = 2/3

Did you mean to say "to a physicist that's good enough right"?

For a Physics exam, I wouldn't actually want to go out on a limb and say they weren't expecting the 0<x<=1 range of values to be included.

But undoubtedly the safest answer is to say the LHS < RHS for 1 < x < sqrt(2), and the RHS is undefined for 0 < x < 1.

Did you mean to say "to a physicist that's good enough right"?

For a Physics exam, I wouldn't actually want to go out on a limb and say they weren't expecting the 0<x<=1 range of values to be included.

But undoubtedly the safest answer is to say the LHS < RHS for 1 < x < sqrt(2), and the RHS is undefined for 0 < x < 1.