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August 25th, 2014, 05:11 AM  #1 
Member Joined: Jun 2014 From: Alberta Posts: 56 Thanks: 2  The Sum of (1/10)^n As n Goes to Infinity
I am using the formula (1  r^(n+1))/(1r) to find the Sum of (1/10)^n as n goes from 1 to infinity. When I plug in the numbers, I get (11/10^(n+1))/(11/10). When n goes to infinity, I get (10)/(11/10) = 10/9. But I don't think it's the right answer. 
August 25th, 2014, 05:59 AM  #2 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus 
$\displaystyle \sum_{k=0}^{\infty}x^k=\frac{1}{1x}$ for $x<1$. $\displaystyle \left\frac{1}{10}\right<1$, so we know that $\displaystyle \sum_{k=0}^{\infty}\left(\frac{1}{10}\right)^{k}= \frac{1}{1\frac{1}{10}}=\frac{1}{\frac{9}{10}}=\frac{10}{9} \Leftrightarrow \sum_{k=1}^{\infty}\left(\frac{1}{10}\right)^k= \frac{10}{9}1= \frac{1}{9}$. 
August 25th, 2014, 04:17 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, Mathbound! Quote:
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: $\:S \;=\;\dfrac{a}{1r}$ The answer is: $\:S \;=\;\dfrac{\frac{1}{10}}{1  \frac{1}{10}} \;=\;\dfrac{\frac{1}{10}}{\frac{9}{10}} \;=\;\dfrac{1}{9}$  
March 30th, 2019, 06:57 PM  #4 
Newbie Joined: Mar 2019 From: Pakistan Posts: 1 Thanks: 0  ...
What if the "n" doesn't starts from 1 it starts from 0 to infinity.

March 30th, 2019, 07:37 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,631 Thanks: 1470  
March 30th, 2019, 11:17 PM  #6 
Senior Member Joined: Aug 2012 Posts: 2,424 Thanks: 759 
Classic "off by one" error, like forgetting that an array starts at 0 in a computer program.


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