# Infinite Geometric Series

#### ToXic01

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:

#### DeanSchlarbaum

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?

#### Anonymous1

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

#### 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

#### Anonymous1

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?

#### ToXic01

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

#### Anonymous1

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

#### Anonymous1

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 $$\displaystyle a=1.$$

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

#### Plato

MHF Helper
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 $$\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}$$