G
Guest
Guest
I'm trying to work out the following:
. . . Find the sum of the infinite series:
. . . . . .\(\displaystyle \L \left( {\frac{4}{5} \right)^3\, -\, \left( {\frac{4}{5} \right)^4\, +\, \left( {\frac{4}{5} \right)^5 \, -\, \left( {\frac{5}{5} \right)^6\, +\, ...\)
As this is an infinite geometric series, with a = 4/5 and r = -(4/5) (right?), I used the equation:
. . .\(\displaystyle \L \sum\limits_{i = 3}^\infty \, {\left( { - \frac{4}{5}} \right)^i }\)
...which gives:
. . .\(\displaystyle \L \frac{\left( {\frac{4}{5}} \right)}{1\, -\, \left(- \frac{4}{5} \right)}\, =\, \frac{4}{9}\)
Please can someone check if this is right?
:?:
. . . Find the sum of the infinite series:
. . . . . .\(\displaystyle \L \left( {\frac{4}{5} \right)^3\, -\, \left( {\frac{4}{5} \right)^4\, +\, \left( {\frac{4}{5} \right)^5 \, -\, \left( {\frac{5}{5} \right)^6\, +\, ...\)
As this is an infinite geometric series, with a = 4/5 and r = -(4/5) (right?), I used the equation:
. . .\(\displaystyle \L \sum\limits_{i = 3}^\infty \, {\left( { - \frac{4}{5}} \right)^i }\)
...which gives:
. . .\(\displaystyle \L \frac{\left( {\frac{4}{5}} \right)}{1\, -\, \left(- \frac{4}{5} \right)}\, =\, \frac{4}{9}\)
Please can someone check if this is right?
:?: