write the sigma notation of (16/81) , (8/27) , (4/9) , ....
Note that $\displaystyle 16/2=8$, $\displaystyle 8/2=4$, ..., and $\displaystyle 81/3=27$, $\displaystyle 27/3=9$, ..., (the terms are in a geometric progression) so the $\displaystyle n$th term may be written $\displaystyle a_n=\frac{16}{2^{n-1}}\frac{3^{n-1}}{81}=\frac{16}{81}\left(\frac32\right)^{n-1}$.
Edit: im sorry, that would make your sum $\displaystyle \sum_{n=0}^\infty\frac{16}{81}\left(\frac32\right) ^n$. i hope that helps.
Hello, lucifer_x!
Write the sigma notation of (16/81) , (8/27) , (4/9) , ....
"Sigma notation" means that the terms are to be added.
So we have: .$\displaystyle \frac{2^4}{3^4} + \frac{2^3}{3^3} + \frac{2^2}{3^2} + \hdots \;=\;\left(\frac{2}{3}\right)^4 + \left(\frac{2}{3}\right)^3 + \left(\frac{2}{3}\right)^2 + \hdots $
Therefore: .$\displaystyle \sum^{\infty}_{n=1} \left(\frac{2}{3}\right)^{5-n} $