Infinite Geometric Series

Mar 2010
27
0
Find the sum of the given infinite geometric series if possible.
The Answer is 0.5. How do I solve this to get to the answer


\(\displaystyle \sum_{i=1}^{\infty} 2(\frac{1}{5})i \)


Thanks
 
Last edited:
Nov 2009
517
130
Big Red, NY
A question why is the answer 0.5

infinity
SIGMA 2(1/5)i
r=1

Thanks
\(\displaystyle \sum_{i=1}^{\infty} 2(\frac{1}{5})^i = 2\left\{[\sum_{i=0}^{\infty} (\frac{1}{5})^i] -1 \right\} = 2(\frac{1}{1-\frac{1}{5}} -1) = \frac{1}{2}\)

aNon1
 
Mar 2010
27
0
Apparently I don't know how to use latex for symbols I had to copy Anonymous1 code to fix the problem. See my op post for the correct question. Thanks
 
Nov 2009
517
130
Big Red, NY
Apparently I don't know how to use latex for symbols I had to copy Anonymous1 code to fix the problem. See my op post for the correct question. Thanks
Do you understand how I arrived at the solution?
 
Mar 2010
27
0
Do you understand how I arrived at the solution?
Because of my incorrect question, the ^i is not suppose to be like that. So I am not sure if you had the correct question for me to understand.However, to answer you question I am still not sure how you came to that conclusion

Thanks
 
Nov 2009
517
130
Big Red, NY
The edited question is not a geometric series, and becomes infinitely large.

\(\displaystyle \sum_{i=1}^{\infty} 2(\frac{1}{5})i = \frac{2}{5}\sum_{i=1}^{\infty}i \to \infty\)

On the other hand...

\(\displaystyle \sum_{i=1}^{\infty} 2(\frac{1}{5})^i\)

...is a geometric series that converges to \(\displaystyle \frac{1}{2}.\)
 
Nov 2009
517
130
Big Red, NY
The geometric series formula is \(\displaystyle \sum_{i=0}^{\infty} a(r)^i = a\frac{1}{1-r}\)

Your sum is:

\(\displaystyle \sum_{i=1}^{\infty} 2(\frac{1}{5})^i =2\sum_{i=1}^{\infty} (\frac{1}{5})^i \)

Notice the indices go from \(\displaystyle i=1\) to \(\displaystyle \infty.\) We want it to go from \(\displaystyle i=0\) to \(\displaystyle \infty\) so that we can use the geometric series formula.

The \(\displaystyle i=0\) term is \(\displaystyle (\frac{1}{5})^0 = 1.\)

So I both added \(\displaystyle 1\) (by changing the sum from \(\displaystyle [1,\infty]\) to \(\displaystyle [0,\infty]\) ) and subtracted \(\displaystyle 1.\)

\(\displaystyle = 2\left\{[\sum_{i=0}^{\infty} (\frac{1}{5})^i] -1 \right\} \)

Now, apply the formula for the geometric series on \(\displaystyle [\sum_{i=0}^{\infty} (\frac{1}{5})^i]\) where \(\displaystyle r=\frac{1}{5}\) and \(\displaystyle a=1.\)

\(\displaystyle = 2(\frac{1}{1-\frac{1}{5}} -1) = \frac{1}{2}\)
 

Plato

MHF Helper
Aug 2006
22,455
8,631
Why not learn to post in symbols? You can use LaTeX tags.
[noparse]\(\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}\)[/noparse] gives \(\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}\)