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
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
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 $\displaystyle \frac{1}{2}.$
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 $\displaystyle a=1.$
$\displaystyle = 2(\frac{1}{1-\frac{1}{5}} -1) = \frac{1}{2}$
Why not learn to post in symbols? You can use LaTeX tags.
[tex] \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}[/tex] gives $\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}$
$\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}}=\frac{2}{5^1}+\frac{2}{5^2}+\frac{2}{5^3}+.... ...................$
here a=2/5 and $\displaystyle r=\frac{2}{5^2}\div\frac{2}{5^1}=\frac{1}{5}$
$\displaystyle \sum\limits_{k = 1}^\infty {\frac{2}{{5^k }}} = \frac{{\frac{2}{5}}}{{1 - \frac{1}{5}}} = \frac{2}{{5 - 1}} = \frac{1}{2}$