The formula for the the sum of infinite terms of a geometric progression is $S_{\infty} = \dfrac{a}{1 - r}$, where $r < 1$. Why is this only defined for $r < 1$? My teacher doesn't know why.
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What happens when you try to sum a geometric progression for which $r \ge 1$? For example for $r=2$, $a=1$: $1+2+4+8+16+.....$. – Zestylemonzi Aug 13 '16 at 09:31
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Unless $a=0$ (meaning every term is 0), if $r \geq 1$ the terms in the series will grow larger and larger (towards ininity), and so the sum will be infinite.
naslundx
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Assume $r\geq 1$. Then $\{r^n\}_{n\in\mathbf N}$ is not a null sequence. By the nth-term test for divergence the series $$\sum_{n=0}^\infty r^n$$ can not converge.
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If $r> 1$, then $\displaystyle\lim_{n\to +\infty} r^n=+\infty$. Since $$\forall n\in\Bbb N, \sum_{k=0}^n r^k= \frac{1-r^{n+1}}{1-r}$$ we deduce that $$\lim_{n\to +\infty} \sum_{k=0}^n r^k=+\infty$$ hence the formula is not correct.
If $r=1$, then $$\forall n\in\Bbb N, \sum_{k=0}^n r^k= \sum_{k=0}^n 1= n+1$$ which again tends to $+\infty$.
paf
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