prove 1 = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 ...?

Question

Haven't done math in years, just currious to see the actual solution worked out. First person to show the actuall proof get's 10 points.

prove that the limit of sum(1/2n) as n approaches infinity equals one.

for n = all integers

Answer 1

Let Sn be the sum of first n terms

S1=1/2=1 - 1/2
S2=3/4=1 - 1/4
S3=7/8=1 - 1/8

or in general

Sn = 1 - 1/2^n

easily proved by math induction, S1 is trivial as a base.

S[n+1] = Sn + 1/2^(n+1)

by induction hypotheis, Sn = (1 - 1/2^n)

S[n+1] = 1 - 1/2^n + 1/2^(n+1))
S[n+1] = 1 - 2/2^(n+1) + 1/2^(n+1)
= 1 - 1/2^(n+1)

the infinite sum is lim(n->infinity) of Sn
lim(n->infinity) 1 - 1/2^n = lim 1 + lim 1/2^n
= 1 + 0 = 1

Answer 2

If you take the first term away, you get

1/4 + 1/8 + 1/16 + 1/32 + 1/64 + ...

which is precisely half of the original series.

Therefore, if X is the sum of the series, we have

X = 1/2 + (X / 2)

Simple algebra shows that X = 1.

Answer 3

First, establish what a geometric series equals

Z = 1 + x + x^2 + ...
-xZ = -x - x^2 - x^3
Z-xZ = Z(1-x) = 1 [this is the sum of the previous two lines, everything but the 1 cancelling out]
Z = 1/(1-x)

Now applying that to below....

? = 1/2 + 1/4 + 1/8 + 1/16 + ....
? = (1/2) + (1/2)^2 + (1/2)^3 + (1/2)^4 + ...
? = 1/2 *(1 + (1/2) + (1/2)^2 + (1/2)^3 + ...)
2*? = 1 + (1/2) + (1/2)^2 + (1/2)^3 + ...
2*? = 1/[1-(1/2)] = 1/(1/2) = 2
? = 1

Answer 4

As the terms 1/2, 1/4, 1/8 ... define a geometric progression with a ratio of r=1/2, it converges (because |r| < 1), the sum is given by the formula r/(1-r), i.e (1/2)/(1-1/2) = (1/2) * 2 = 1.

Answer 5

A series of the form

a+ar+ar^2+ar^3+...
=\sum_{k=0}^{\infty} ar^k

is called a geometric series.

Using calculus, we can prove that the geometric series diverges if |r| >=1 and converges to a/(1-r) if |r|<1.

In your case, you have a geometric series with a=1/2 and r=1/2. Because |r|=1/2<1, it converges to

a/(1-r)=(1/2)/(1-1/2)
=(1/2)/(1/2)=1.

Note: In your comment, you wrote \sum_{n=1}^{\infty} 1/(2n). This series can be written as 1/2 \sum_{n=1}^{\infty} 1/n. The series \sum_{n=1}^{\infty} 1/n is also a special series called the harmonic series. One reason the harmonic series is interesting is because it diverges: \sum_{n=1}^{\infty} 1/n=+ \infty. Thus, the harmonic series is an example of a series that diverges even though the limit of its terms is 0.

Answer 6

You have a geometic series with a1 = 1/2, and r = 1/2.

The sum of a geometic series is

a1 * (1 - r^n) / (1 - r)

Here n = Infinity, so r^n = 0. You have

the RHS = 1/2 * (1 / (1/2)) = 1.

Answer 7

simple

let S be the sum of the series, then

let S= 1/2+1/4+1/8+.............................. equation (1 )

the S/2 = 0+1/4+ 1/8+................................ equation(2)

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SUBTRUCT S/2 = 1/2 + 0 +0+.................... EQUATION 3

then you have from EQUATION 3: S/2 = I/2 => S = 1

the guys above me making it too complicated without reason!!!