the sum of 1/2+1/4+1/8+1/16-----------equals1.what is sum of 1/2+1/16+1/64+1/256-----------?

Answer 1

The first series is called a geometric series. Using common notation that you would find in a calculus book, it is geometric with a=1/2 and r=1/2 so it converges to a/(1-r)=1.

The second series is more interesting. Because we are dealing with positive term series and the terms of the second series are less than or equal to the corresponding terms of the first series, which converges, the second series converges by the Basic Comparison test (or Comparison test). Using commercial software, I cannot find the exact value of the sum, but the following is the value of the sum out to about 25 decimal places:

N[Sum[1/2^(2^n),{n,0,Infinity}],25]

0.816421509021893143708080

Answer 2

These are infinite series that gradually converge to a final value if an infinite number of terms are included.

These two series converge to different values:

a) 1/2 + 1/4 + 1/8 + 1/16 ... = sum(1/2^n) = 1.000000

b) 1/2 + 1/16 + 1/64 + 1/256... = sum[1/2^(2*n)] = 0.583333

where 'n' ranges from 0 to infinity.

You can try this with a spreadsheet. Create formulas that generate several numbers in the series and add them up. If you plot the cumulative sum as a function of the number of terms, you'll see the asymptotic convergence to a final value.

Answer 3

Buddy I don't think that these two series are equal.The first series is:
(1/2)/(1-(1/2)).....(infinite G.P. series with common ratio 1/2) results in 1.The second one however is a bit different 1/2 is out of the G.P. series with common ratio 1/4 so the result is:(1/16)/(1-(1/4))
result is (1/2+1/16) and that's not unity.so the two series are different.

Answer 4

for geometric countless series with a as first volume and ratio as r, the position r

Answer 5

1/2+1/4+1/8+1/16+.......
= 1/2*(1-1/2^n)/(1-1/2)
=1

1/2+1/16+1/64+1/256+.......
=1/2+1/16*(1-1/4^n)/(1-1/4)
=1/2+1/12
=7/12
=0.58333.....

Answer 6

first is series of
1/2+a+b+c+...
other is
1/2+a^2+b^2+c^2...
they are not equal

Answer 7

Both series approach a similar solution--the number one. Although they are infinitesimally equal, they are never the same. However, for the sake of argument, infinitely close to one is equal to infinitely close to one.

Answer 8

15/16
149/256

Answer 9

there is no exact answer...
but i know u knew it because your the one who
asked this questions

Answer 10

Please re-write your question. Make your given and what you want to do clear.