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$\sum_{n=1}^\infty \frac1{2^n} $

Somehow this gives $a = \frac12$, $r = \frac12$ to put into the formula $\sum_{n=1}^\infty ar^{n-1}$.

whats a general method (if any) for finding $a$ and $r$ in the geometric series formula?

i know that if you put a series in the form $\sum_{n=1}^\infty ar^{n-1} $ you can find if it is divergent (if $|r| \ge 1$) or convergent (if $|r| < 1$) with the sum being $\frac{a}{1-r}$. I am not quite sure however how to obtain $a$ and $r$ from a series and put it into the general form of the geometric series.

my question is HOW do i find the values $a$ and $r$

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  • $\begingroup$ Please make the question clear. $\endgroup$ – jdoicj Dec 11 '13 at 13:18
  • $\begingroup$ im not sure how to make it more clear... the formula is summation[a*r^(n-1)] and a = 1/2, r = 1/2, how can I get a and r to be those values from my original equation Summation[ 1/(2^n)] ? $\endgroup$ – 2316354654 Dec 11 '13 at 13:21
  • $\begingroup$ Go through this meta.math.stackexchange.com/questions/5020/… $\endgroup$ – jdoicj Dec 11 '13 at 13:23
  • $\begingroup$ ok, thank you. . $\endgroup$ – 2316354654 Dec 11 '13 at 13:26
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    $\begingroup$ Question is not clear $\endgroup$ – Way to infinity Dec 11 '13 at 13:40
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We know, $\displaystyle\sum_{k=1}^{n} ar^{k-1}= a\dfrac{1-r^{n}}{1-r}$.

Now, $\displaystyle\sum_{k=1}^\infty \dfrac1{2^k} =\lim_{n\to\infty}\displaystyle\sum_{k=1}^n \dfrac1{2^k} =\lim_{n\to\infty}\displaystyle\sum_{k=1}^n \left(\dfrac12\right)\cdot{\left(\dfrac1{2}\right)}^{k-1}$.

This is in the form of the above formula with $a=\dfrac12$ and $r=\dfrac12$. So the expression becomes,

$\lim_{n\to\infty} \dfrac12\dfrac{1-\left(\frac12\right)^{n}}{1-\frac12} =\lim_{n\to\infty} 1-\left(\frac12\right)^n=1.$


If we instead know, $\displaystyle\sum_{k=1}^{\infty} ar^{k-1}=\dfrac{a}{1-r}$ for $|r|<1$,

Then again,

$\displaystyle\sum_{k=1}^\infty \dfrac1{2^k}=\displaystyle\sum_{k=1}^\infty \left(\dfrac12\right)\left(\dfrac12\right)^{k-1}$

which is in the form of the preceding formula with $a=\frac12$ and $r=\frac12$ (and $|r|<1$).

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  • $\begingroup$ yes, i did not know what "we know" in the first line... $\endgroup$ – 2316354654 Dec 11 '13 at 14:00
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    $\begingroup$ @mathisatroll That's the formula for the sum of a geometric series. $\endgroup$ – Alraxite Dec 11 '13 at 14:01
  • $\begingroup$ actually im still not quite clear on how you acquire 1/2 for a and r in the first place. $\endgroup$ – 2316354654 Dec 11 '13 at 14:07
  • $\begingroup$ i knew the formula as the left hand side of the first line, but no where in the book did it say anything about the right hand side. $\endgroup$ – 2316354654 Dec 11 '13 at 14:08
  • $\begingroup$ @mathisatroll Could you tell me precisely what you know of the sum of a geometric series? I'm not sure where your confusion lies. $\endgroup$ – Alraxite Dec 11 '13 at 14:14
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The number $r$ is the ratio of two consecutive terms in $\sum_{n=1}^\infty ar^{n-1}$. That is, we have $$r=\frac{ar^n}{ar^{n-1}}.$$ And when you have $r$, you should be able to get $a$.

NOTE: If you have a series $\sum b_n$ and the ratio $\frac{b_n}{b_{n-1}}$ depends on $n$, then you know that your series is NOT geometric.

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  • $\begingroup$ im still confused on how you get "r" to begin with.. seeing as your formula has "r" = (an expression with r in it as well). i tried taking the ratio of the first two terms of the series 1/(2^n) but that left me with 2, not 1/2.. $\endgroup$ – 2316354654 Dec 12 '13 at 7:11
  • $\begingroup$ @mathisatroll you have to take the ratio in the right order: $(1/2^n)/(1/2^{n-1})=1/2$. $\endgroup$ – M Turgeon Dec 12 '13 at 13:44
  • $\begingroup$ i just wrote that below 4 hours ago. $\endgroup$ – 2316354654 Dec 12 '13 at 15:24
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well this is the simple answer to the simple question.

it appears from further research that $a$ is simply the first evaluation of the series and $r$ is $\frac{a_2}{a_1}$. which makes $a_1 = \frac12, a_2 = \frac14,$ so $\frac{a_2}{a_1} = \frac12$, therefor $r = \frac12$.

nowwwww I see, I was dividing $a_1$ by $a_2$, not the other way around, to get $2$ earlier.

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  • $\begingroup$ Please use the edit link on your question to add additional information. The Post Answer button should be used only for complete answers to the question. $\endgroup$ – mau Dec 12 '13 at 10:20
  • $\begingroup$ but this WAS the complete answer to my actual QUESTION $\endgroup$ – 2316354654 Dec 12 '13 at 11:04

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