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What observation can make about the successive partial sums of this sequence? In particular, what number does it appear that the sum will always be smaller than?
Answer:
Terms Sum

1
1, 1/3
1,1/3,1/9
1,1/3,1/9,1/27

2007-06-29 18:50:55 · 8 answers · asked by Sheri C 1 in Science & Mathematics Mathematics

8 answers

The partial sum of the sequence generated by 3^(-n) incrementing by n=1 at every step converges asymptotically on 1.5, each new term failing to bridge the gap between the accumulative total and 1.5 by ever smaller amounts, but failing nevertheless to bridge it. This is best illustrated by a table:

Further Terms and Partial Sums

1 and 1
1/3 and 1 1/3
1/9 and 1 4/9
1/27 and 1 13/27
1/81 and 1 40/81
1/243 and 1 121/243
1/729 and 1 364/729
1/2187 and 1 1093/2187
etc

The numerator in the partial sum is simply the accumulative total of the denominators of all the terms to date, each new numerator being (three times the previous one) plus one.

The corresponding sequence 1, 1/2, 1/4, 1/8, 1/16, etc similarly converges asymptotically on 2.

2007-06-29 19:02:02 · answer #1 · answered by Anonymous · 0 0

1
1 + 1/3 = (3 + 1)/3 = 4/3 = 3/2 - 1/6 ≈ 1.3333333333333333333333333333333
1 + 1/3 + 1/9 = (9 + 3 + 1)/9 = 13/9 = 3/2 - 1/18 ≈ 1.4444444444444444444444444444444
1 + 1/3 + 1/9 + 1/27 = (27 + 9 + 3 + 1)/27 = 40/27 = 3/2 - 1/54 ≈ 1.4814814814814814814814814814815
1 + 1/3 + 1/9 + 1/27 + 1/81 = (81 + 27 + 9 + 3 + 1)/81 = 121/80 = 3/2 - 1/162 ≈ 1.4938271604938271604938271604938
1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 = (243 + 81 + 27 + 9 + 3 + 1)/243 = 3264/243 = 3/2 - 1/486 ≈ 1.4979423868312757201646090534979
1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729= (729 + 243 + 81 + 27 + 9 + 3 + 1)/729 = 1,093/729 = 3/2 - 1/1458 ≈ 1.4993141289437585733882030178326
It sure looks like the sum will always be smaller than 3/2 or 1.5

2007-06-30 02:55:48 · answer #2 · answered by Helmut 7 · 0 0

the bottom number of the fractions are multiplied three to get the next number
ie.
1/1 x 1/3 = 1/3
1/3 x 1/3 = 1/9
1/9 x 1/3 = 1/27

2007-06-30 02:01:18 · answer #3 · answered by Anonymous · 0 0

The answer 1.5 is correct. There is a simple way to do these geometric series sums:

x=1+ 1/3 + 1/9 +...
3x = 3 + 1 + 1/3 + 1/9 +...

So:
3x = x+3
2x=3
x=1.5

2007-06-30 02:09:16 · answer #4 · answered by Thomas M 6 · 2 0

Does 1/81 or 1/243 make the sum greater than 1.5? Try it.

2007-06-30 01:58:21 · answer #5 · answered by novangelis 7 · 0 0

sum = 3/2

2007-06-30 02:02:52 · answer #6 · answered by CPUcate 6 · 0 0

S∞ = a / (1 - r) where "a" is first term and r is common ratio.
a = 1
r = 1/3
S∞ = 1 / ( 1 - (2/3))
S∞ = 1 / (1/3)
S∞ = 3
Sum will always be less than 3.

2007-06-30 11:36:20 · answer #7 · answered by Como 7 · 0 0

1/3^0 = 1
1/3^1 = 1/3
1/3^2 = 1/9
1/3^3 = 1/27
1/3^4 = 1/81
.
.
.
1/3^n = 1/3^n

2007-06-30 01:58:01 · answer #8 · answered by jimschem 4 · 0 0

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