1. ## Algebra summation

t_i=(3/u)^(i)*(1-3/u)

What's the summation (t_i) from 3 to infinity?

I know the answer is (3/u)^3 but don't know how to get it.

I would appreciate your help. Thank you!

2. ## Re: Algebra summation

\displaystyle \begin{align*} \sum_{i = 3}^{\infty}{ \left[ \left( \frac{3}{u} \right) ^i \,\left( 1 - \frac{3}{u} \right) \right] } &= \sum_{ i = 3}^{\infty}{ \left[ \left( \frac{3}{u} \right) ^i - \left( \frac{3}{u} \right) ^{i + 1} \right] } \\ &= \sum_{i = 3}^{\infty}{ \left( \frac{3}{u} \right) ^i } - \sum_{i = 3}^{\infty}{ \left( \frac{3}{u} \right) ^{i + 1} } \\ &= \sum_{i = 3}^{\infty}{ \left( \frac{3}{u} \right) ^i } - \sum_{i = 4}^{\infty}{ \left( \frac{3}{u} \right) ^i } \\ &= \frac{\left( \frac{3}{u} \right) ^3}{1 - \frac{3}{u} } - \frac{\left( \frac{3}{u} \right) ^4}{1 - \frac{3}{u}} \textrm{ provided } \left| \frac{3}{u} \right| < 1 \\ &= \frac{\left( \frac{3}{u} \right) ^3 - \left( \frac{3}{u} \right) ^4}{1 - \frac{3}{u}} \\ &= \frac{\left( \frac{3}{u} \right) ^3 \,\left( 1 - \frac{3}{u} \right) }{1 - \frac{3}{u}} \\ &= \left( \frac{3}{u} \right) ^3 \end{align*}

3. ## Re: Algebra summation

Originally Posted by mathfn
t_i=(3/u)^(i)*(1-3/u)
What's the summation (t_i) from 3 to infinity?
I know the answer is (3/u)^3 but don't know how to get it.
Here are some hints that might simply things.
If $\bf{|r|<1}$ and $J\in\mathbb{Z}$ then $\displaystyle\sum\limits_{k = J}^\infty {c{r^k}} = c\frac{{{r^J}}}{{1 - r}}$

With that in mind: If $\left|{\left( {\dfrac{3}{u}} \right)}\right|<1$ then $\displaystyle\sum\limits_{k = 3}^\infty {\left( {1 - \frac{3}{u}} \right){{\left( {\frac{3}{u}} \right)}^k}} = \left( {1 - \frac{3}{u}} \right)\dfrac{{{{\left( {\frac{3}{u}} \right)}^3}}}{{1 - \frac{3}{u}}}={\left( {\dfrac{3}{u}} \right)}^3$