1
$\begingroup$

This problem comes from Solow's book, 2nd edition.

What is wrong with the proof given below?

If $r$ is a real number with $|r| \leq 1$, then for all integers $n \geq 1, 1 + r + r^2 + \ldots + r^{n-1} = \dfrac{1-r^n}{1-r}$.

Proof. This statement is clearly true for $n=1$. Assume it is true for $n$. Then, for $n+1$, one has

$1 + \ldots + r^n = \dfrac{1-r^n}{1-r} + r^n$ $= \dfrac{1 - r^n + r^n - r^{n+1}}{1-r}$ $= \dfrac{1-r^{n+1}}{1-r}$

// The only issue I see with this is that the case for $r=1$ is untrue (as LHS is positive and RHS is undefined). Are there any bigger issues that I'm missing?

$\endgroup$
1
  • 1
    $\begingroup$ That's the only issue I see. The formula should be valid only for $\mid r \mid \lt 1$, not $\le$. $\endgroup$
    – apnorton
    Jan 18, 2014 at 3:33

2 Answers 2

1
$\begingroup$

The statement which is true for every $r$ is completely algebraic and reads $$ (1-r)(1+r+\cdots+r^{n-1})=1-r^n. $$ (Proof: Let $s=1+r+\cdots+r^{n-1}$ then $rs=r+r^2+\cdots+r^{n}=s+r^n-1$ hence $(1-r)s=s-rs=1-r^n$.)

The expression in the post follows as soon as one can divide by $1-r$, for example, for every real number $r\ne1$.

$\endgroup$
0
$\begingroup$

Yes, you are right. It clearly makes no sense for r=1.

And this proposition is true for all r≠1

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .