What is the sum of the infinate geometric series: 64,32,16,8,4,2,1......?

Suppose S = 64 + 32 + 16 + 8 + ...

So 2S = 128 + 64 + 32 + 16 + ...

Subtract those:

2S - S = 128

S = 128

Sum = a/(1 - r)

a = 64 and r = 1/2

Sum = (64)/(1/2) = 128

let sum of the series be S

S=64+32+16+8+....upto infinite

So sum of the series will be

S=a/(1-r)

Here a=first tirm of the series and

r=commom ratio of GP

a=64 & r=1/2

On putting the values you will get

S=128

128

sum(64 * (1/2)^n , n = 0 , n = inf) =

64 * sum((1/2)^n , n = 0 , n = inf)

sum((1/2)^n , n = 0 , n = t) = S

S = (1/2)^0 + (1/2)^1 + (1/2)^2 + .... + (1/2)^t

S * (1/2) = (1/2) + (1/2)^2 + .... + (1/2)^t + (1/2)^(t + 1)

S * (1/2) - S = (1/2) * (1/2)^t - 1

S * (1/2 - 1) = (1/2) * (1/2)^t - 1

S * (-1/2) = (1/2) * (1/2)^t - 1

S * (1/2) = 1 - (1/2) * (1/2)^t

S = 2 - (1/2)^t

64 * S =

64 * (2 - (1/2)^t)

t = inf

64 * (2 - (1/2)^inf) =

64 * (2 - 0) =

64 * 2 =

128