Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm in Precalc 2, and the question being asked is: "Show that the sum of the following infinite geometric series is $\dfrac{3}{2} = \dfrac{\sqrt{3}}{\sqrt{3} + 1} + \dfrac{\sqrt{3}}{\sqrt{3} + 3} \cdots $

I know that Infinite Series $S=\dfrac{a}{1-r}$, my problem is knowing what my $a$ and $r$ are. I'm pretty sure $a = \sqrt{3}$, but $r$ is what's confusing.

share|cite|improve this question
up vote 4 down vote accepted

We have $a=\dfrac{\sqrt{3}}{\sqrt{3}+1}$ and $r=\dfrac{1}{\sqrt{3}}$.

The $a$ should take no thinking. In the formula you quoted, $a$ is always the first term of the geometric series.

For the common ratio $r$, divide the second term by the first term. We get after minor cancellation that $r$ is $\frac{\sqrt{3}+1}{3+\sqrt{3}}$. One could leave it that way, or simplify by noting that $3+\sqrt{3}=\sqrt{3}(\sqrt{3}+1)$.

share|cite|improve this answer
As soon as I read that second line, I proceeded with a massive facepalm. Thanks so much! – PJ Johnson May 6 '14 at 5:51
You are welcome. The basic simplicity was hidden behind a bunch of square roots. – André Nicolas May 6 '14 at 5:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.