$\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 $r$

  • $\begingroup$ 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
  • 1
    $\begingroup$ Question is not clear $\endgroup$ – Way to infinity Dec 11 '13 at 13:40

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 $|r|<1$).

  • $\begingroup$ yes, i did not know what "we know" in the first line... $\endgroup$ – 2316354654 Dec 11 '13 at 14:00
  • 1
    $\begingroup$ @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

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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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