The sum of n g.p. terms with first term a and common ratio r is given by $$ S_{n} = a\cdot\frac{1 - r^n}{1 -r} provided \ r \not= 1 $$ But I'm confused as to what happen when r $ \lt 1$. My module says that $r^n$ decreases as n increases. Thus when n becomes very large that is as $ n \to \infty $, $r \to 0 $. But let's suppose if $r = -2$ then $r^n$ changes to positive and negative depending whether n is even or odd. Then how does one get this "as $ n \to \infty $, $r \to 0 $."

  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – lab bhattacharjee Jan 27 '14 at 14:03
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    $\begingroup$ It isn't $\;r<1\;$ but $\;|r|<1\iff -1<r<1\;$...! $\endgroup$ – DonAntonio Jan 27 '14 at 14:05
  • $\begingroup$ If $|r|<1$, then $|r^n|\rightarrow 0$. If $|r|>1$, then $|r^n|\rightarrow\infty$ (then $\lim_{n\rightarrow\infty} S_n$ does not exist). $\endgroup$ – David Mitra Jan 27 '14 at 14:05

Note that $r^n$ converges to $ 0$ not when $$r<1,$$ but rather when $$|r|<1.$$ That is, only for $-1 < r < 1$, not for $r < -1$ and in particular not for $r = -2$.

For $r < -1$, the values of $r^n$ diverge as $n$ increases, just as for $r>1$. The finite sum formula still holds: the sum of the first $n$ terms of the series is still $S_n = a\frac{1-r^n}{1-r}$. But since $r^n$ diverges, so does $S_n$.


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