1. ## Infinite Geometric Series

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

2. i can't understand your Que

3. Originally Posted by ToXic01
A question why is the answer 0.5

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

Thanks
ToXic01: I cannot understand either. Can you clarify?

4. Originally Posted by ToXic01
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

5. 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

6. Originally Posted by ToXic01
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?

7. Originally Posted by Anonymous1
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

8. 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 $\displaystyle \frac{1}{2}.$

9. The geometric series formula is $\displaystyle \sum_{i=0}^{\infty} a(r)^i = a\frac{1}{1-r}$

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

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 $\displaystyle a=1.$

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

10. Why not learn to post in symbols? You can use LaTeX tags.
$$\sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}$$ gives $\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}$

11. yea i doubled checked

12. can i ask how you got the ratio 1/5

13. $\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}}=\frac{2}{5^1}+\frac{2}{5^2}+\frac{2}{5^3}+.... ...................$
here a=2/5 and $\displaystyle r=\frac{2}{5^2}\div\frac{2}{5^1}=\frac{1}{5}$
$\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}$