Thread: Geometric infinite series

I am confused on just one review question... involving geometric infinite series. E = sigma notation.

E(k=0 to infinity) [2(1/4)^k + 3(-1/5)^k]

Thank you!

okay so I was thinking seperate the two so E(k=0 to infinity) 2(1/4)^k + E(k=0 to inf) 3(-1/5)^k

Could I pull in the 2 and 3 and make it (1/2) and (-3/5) respectively?

Originally Posted by pakman

I am confused on just one review question... involving geometric infinite series. E = sigma notation.

E(k=0 to infinity) [2(1/4)^k + 3(-1/5)^k]

Thank you!

Since both series are congergent you can do each one seperately.

First (1/4)^k is as a geometric sum: 1/(1-1/4)=4/3
Multiplied by a factor of two infront of it: 8/3

The second (-1/5)^5 is as a geometric sum: 1/(1+1/5)=5/6
Multiplied by a factor of 3 is: 3/2
When added yiedls,
3/2+8/3

Originally Posted by ThePerfectHacker

Since both series are congergent you can do each one seperately.

First (1/4)^k is as a geometric sum: 1/(1-1/4)=4/3
Multiplied by a factor of two infront of it: 8/3

The second (-1/5)^5 is as a geometric sum: 1/(1+1/5)=5/6
Multiplied by a factor of 3 is: 3/2
When added yiedls,
3/2+8/3

Thank you