Calculate the sum for the infinite geometric series

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Discussion Overview

The discussion revolves around calculating the sum of the infinite geometric series represented by the terms $4+2+1+\frac{1}{2}+...$. Participants explore the formula for the sum of a geometric series and the values of the first term and common ratio.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant identifies the common ratio as $\frac{1}{2}$ and suggests using the formula for the sum of an infinite geometric series, $\displaystyle\sum_{n=0}^{\infty} a r^n = a \left(\frac{1}{1-r}\right)$.
  • Multiple participants calculate the sum using the first term $a=4$ and confirm the result as $8$ based on the formula.
  • Another participant incorrectly states the first term as $a=5$, but this is corrected by a later reply asserting that $a=4$ is the correct value.
  • There is a repetition of the calculation steps by different participants, reinforcing the same result of $8$.

Areas of Agreement / Disagreement

Participants generally agree on the calculation method and the result of $8$, but there is a disagreement regarding the value of the first term, with one participant mistakenly stating it as $5$ before being corrected.

Contextual Notes

The discussion does not resolve the initial confusion regarding the first term, as it is stated incorrectly by one participant before being corrected by another.

karush
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Calculate the sum for the infinite geometric series
$4+2+1+\frac{1}{2}+...$

all I know is the ratio is $\frac{1}{2}$

$\displaystyle\sum_{n}^{\infty}a{r}^{n}$
assume this is used
 
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karush said:
Calculate the sum for the infinite geometric series
$4+2+1+\frac{1}{2}+...$

all I know is the ratio is $\frac{1}{2}$

$\displaystyle\sum_{n}^{\infty}a{r}^{n}$
assume this is used

what is sum of Geometric series$\displaystyle\sum_{n= 0}^{\infty}a{r}^{n}= a (\frac{1}{1-r})$
using above
$= 4 (\frac{1}{1-\frac{1}{2}})= 4 * 2 = 8$
 
Calculate the sum for the infinite geometric series
$4+2+1+\frac{1}{2}+...$

$a=4 \ \ r=\frac{1}{2}$

$\displaystyle\sum_{n= 0}^{\infty}a{r}^{n}= a (\frac{1}{1-r})
= 4 (\frac{1}{1-\frac{1}{2}})= 4 * 2 = 8$
 
karush said:
Calculate the sum for the infinite geometric series
$4+2+1+\frac{1}{2}+...$

$a=5 \ \ r=\frac{1}{2}$

$\displaystyle\sum_{n= 0}^{\infty}a{r}^{n}= a (\frac{1}{1-r})
= 4 (\frac{1}{1-\frac{1}{2}})= 4 * 2 = 8$

$a = 4$ and not 5
 
Saw it, got it
 

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