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AdrianCavinder · Apr 16, 2012, 11:33 PM

I am really stumped by this question. It’s in the AS level course (11th grade).
An infinite geometric series has first term a and sum to infinity b, where b ≠ 0. Prove that a lies between 0 and 2b.
How do I go about proving this? I can find the sum to infinity in terms of a, b and r, but not sure if I’m heading in the right direction. Any help would be appreciated.

jcaron2 · May 8, 2012, 07:59 PM

Originally Posted by AdrianCavinder

I am really stumped by this question. It’s in the AS level course (11th grade).
An infinite geometric series has first term a and sum to infinity b, where b ≠ 0. Prove that a lies between 0 and 2b.
How do I go about proving this? I can find the sum to infinity in terms of a, b and r, but not sure if I’m heading in the right direction. Any help would be appreciated.

$\sum_{n=0}^\infty ar^n = b$

Since the series converges, you know that r must lie between -1 and 1 (that's a precondition of convergence for a geometric series).

Given this condition, we know that the series converges to:

$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}$

Thus, we can say

$\frac{a}{1-r}=b$

or

$a=(1-r) \cdot b$ ,

where $-1 \leq r \leq 1$

Plugging in the two extremes for r (-1 and 1), you'll see that the possible values for a are bounded by 0 and 2b, with intermediate values of r falling somewhere in between.