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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.

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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 $3+\sqrt{3}=\sqrt{3}(\sqrt{3}+1)$.

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

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