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August 25th, 2014, 04:11 AM   #1
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The Sum of (1/10)^n As n Goes to Infinity

I am using the formula (1 - r^(n+1))/(1-r) to find the Sum of (1/10)^n as n goes from 1 to infinity.

When I plug in the numbers, I get (1-1/10^(n+1))/(1-1/10).

When n goes to infinity, I get (1-0)/(1-1/10) = 10/9. But I don't think it's the right answer.
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August 25th, 2014, 04:59 AM   #2
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Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus
$\displaystyle \sum_{k=0}^{\infty}x^k=\frac{1}{1-x}$ for $|x|<1$.

$\displaystyle \left|\frac{1}{10}\right|<1$, so we know that $\displaystyle \sum_{k=0}^{\infty}\left(\frac{1}{10}\right)^{k}= \frac{1}{1-\frac{1}{10}}=\frac{1}{\frac{9}{10}}=\frac{10}{9} \Leftrightarrow \sum_{k=1}^{\infty}\left(\frac{1}{10}\right)^k= \frac{10}{9}-1= \frac{1}{9}$.
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August 25th, 2014, 03:17 PM   #3
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Hello, Mathbound!

Quote:
I am using the formula (1 - r^(n+1))/(1-r)$\quad$ This is incorrect!
to find the Sum of (1/10)^n as n goes from 1 to infinity.

When I plug in the numbers, I get (1-1/10^(n+1))/(1-1/10).

When n goes to infinity, I get (1-0)/(1-1/10) = 10/9.
But I don't think it's the right answer.

You have: $\displaystyle\;S \;=\;\sum^{\infty}_{n=1}\left(\frac{1}{10}\right)^ n \;=\;\frac{1}{10} + \left(\frac{1}{10}\right)^2 + \left(\frac{1}{10}\right)^3 + \cdots $

You forgot the first term $a \,=\,\frac{1}{10}$.

The formula is: $\:S \;=\;\dfrac{a}{1-r}$

The answer is: $\:S \;=\;\dfrac{\frac{1}{10}}{1 - \frac{1}{10}} \;=\;\dfrac{\frac{1}{10}}{\frac{9}{10}} \;=\;\dfrac{1}{9}$

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March 30th, 2019, 05:57 PM   #4
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What if the "n" doesn't starts from 1 it starts from 0 to infinity.
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March 30th, 2019, 06:37 PM   #5
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Quote:
Originally Posted by Kvs View Post
What if the "n" doesn't starts from 1 it starts from 0 to infinity.
add $\left(\dfrac{1}{10}\right)^0 = 1$
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March 30th, 2019, 10:17 PM   #6
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Classic "off by one" error, like forgetting that an array starts at 0 in a computer program.
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