sum = 3^(-1) + 3^(-2)+ 3^(-3)+ 3^(-4)+ . . . .?

Answer 1

if its upto infinity,then the abov bcoms an infinite GP
so answer is:(1/3)/1-(1/3)=1/2

Answer 2

Hey there!

The sum of 3^(-1) + 3^(-2)+ 3^(-3)+ 3^(-4)+ . . . , can be rewritten as (1/3)+(1/3)^2+(1/3)^3+(1/3)^4+. . .

This sum is a geometric series, with the common ratio, r, being 1/3 and the first term, a1, being 1/3. (I can't write in subscripts).

The sum for infinite geometric series is a1/(1-r).

Here's the answer.

s=a1/(1-r) --> Write the formula for the sum of infinite geometric series.
s=1/3/(1-1/3) --> Substitute 1/3 for a1 and 1/3 for r.
s=1/3/2/3 --> Subtract 1 and 1/3.
s=1/2 Divide 1/3 and 2/3.

So the sum is 1/2.

Here's a useful trick.

For the infinite sum, (1/t)^n, the sum of that geometric series is 1/n-1. This formula is directly proved from the above formula.

So the answer is 1/2.

Hope it helps!

Answer 3

Sum of an infinite series:
S = a/(1 - r)

Where a = the first term, and
r = the ratio

The ratio is r = 1/3, and the first term is 3^(-1), or 1/3. Therefore, the sum is

S = (1/3)/[ (1 - 1/3) ]
S = (1/3)/[ 2/3 ]
S = 1/2

Answer 4

S = 1/3 + 1/9 + 1/27 + 1/81 + ...
Multiply through by 3 :
3S = 1 + 1/3 + 1/9 + 1/27 + ...

But this is just :
3S = 1 + S
(Infinity does strange things)

Now subtract S from both sides :
2S = 1

Divide both sides by 2 :
S = 1/2

Answer 5

Geometric series
a = 1/3 , r = 1/3
S∞ = a / (1 - r)
S∞ = (1/3) / (2/3)
S∞ = 1 / 2

Answer 6

sum = 1/ 3^n
where n = a positive integer like 1,2,3,4,5,6,7, ..