MHB Calculate the sum for the infinite geometric series

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The infinite geometric series discussed is 4 + 2 + 1 + 1/2 + ..., with the first term (a) being 4 and the common ratio (r) being 1/2. The formula for the sum of an infinite geometric series is S = a / (1 - r). Applying this formula, the sum is calculated as S = 4 / (1 - 1/2), which simplifies to S = 4 * 2 = 8. There was a brief confusion regarding the value of a being 5, but it was confirmed that a is indeed 4. The final sum of the series is 8.
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
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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