Find the sum of 1/4+1/16+1/64...?

Question

*This is a series, and goes on for infinity*

Answer 1

the ratio is 1/4

sum = a1 / (1 - r)

sum = 1/4 / (1 - 1/4)

sum = 1/3

Answer 2

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Answer 3

We can use the geometric series Summation from n=1 to infinity of ar^(n-1)=a+ar+ar^2... where a=1 and r=1/4. This would give 1+1/4+(1/4)^2+(1/4)^3+..., which is close to your sum.

The sum equals a/(1-r), so the answer would be 1/(1-1/4) or 4/3.

BUT, since the summation I described above is 1+1/4+1/16+1/64+... and your summation is 1/4+1/16+1/64+... , you must subtract 1 from 4/3, thus giving you the correct answer of 1/3.

Answer 4

This is an infinite geometric series which has a sum if and only if the common ratio is betwen -1 and 1. In this case, r = 1/4, so we say the sum converges to:

a1 / (1-r)
where a1 is the first term and r is the common ratio

So the inifinite sum = 1/4 / (1-1/4) = (1/4) / (3/4)= 1/3

Answer 5

Let S = 1/4 + 1/16 + 1/64 + ......... infinity.
Multiplying on both sides with 1/4 ;
S/4 = 1/16 + 1/64 + 1/256 + .......... infinity.

Subtracting the second relation from the first;
S - S/4 = {1/4 + 1/16 +1/64 + 1/256 + .......} - {1/16 + 1/64 + 1/256 + ........ }
=> 3S/4 = 1/4
=> S = 1/3

Answer 6

It's a geometric series with a = r = 1/4
Answer = 1/4 / (1- 1/4) = 1/3

Answer 7

This is a geometric series:

a = 1/4
r = 1/4

The sum to infinity is a/(1 - r)

s = (1/4)/(3/4) = 1/3

Answer 8

Common ratio , r = 1/4
First term is a = 1/4
S∞ = a / (1 - r) = 1/4 / (3/4) = 1/3

Answer 9

(1 / 4) + (1 / 16) + (1 / 64) = 0.328125

Answer 10

16/64+4/64+1/64=21/64 or .328125
If it goes on for ever there is no answer.