# Limit (infinity)

1. Apr 8, 2009

### Dell

how do i find the limit for the following , where n->infinity

1/2 + 3/4 + 5/8 + 7/16 +9/32 +......

i see that the numerator starts at 1 and has jumps of +2, giving me all the odd numbers

the denominator starts at 2 with jumps of *2 giving all the powers of 2

so i have.... + (2n-1)/2^n

but how do i find the sum of the series? where n=infinity

??

if i had one fraction, (1+3+5+7+9....)/(2+4+8+16....) then i know i could use the equations for sum of a series, but how do i dela with each one as its own fraction.

i am looking for the lim of the sum, not the lim of (2n-1)/2^n

lim$$\sum$$(2n-1)/2^n

2. Apr 8, 2009

### Gib Z

Well, you did the first step correctly, in finding the general term. Now, if we split the numerator, the second term is just a geometric series, whilst the first term, $n/2^{n-1}$ is not quite as easy. To do this, we must use the result

$$\sum_{n=0}^{\infty} nr^{n-1} = \frac{1}{(r-1)^2}$$. Play around with it, see what you can substitute to get it to fit what you have.

3. Apr 8, 2009

### Dell

i took a bit of a different route, getting there but im stuck. maybe you can help me with my way before i try yours

the sum S is

S = 1/2 + 3/4 + 5/8 + 7/16 +9/32 +...... (2n-1)/2^n

now i take double that and i get
2S = 1 + 3/2 + 5/4 + 7/8 + 9/16 +......(4n-2)/2^n

now if i subtract 2S-S i get
2S = 1 + 3/2 + 5/4 + 7/8 + 9/16 +......(4n-2)/2^n
S =//// 1/2 + 3/4 + 5/8 + 7/16 +9/32 +...... (2n-1)/2^n

as you see, if i subtract S from 2S all the middle fractions have a matching one (eg, 3/2 -1/2 =2/2 5/4 -3/4=2/4 etc) only the 1 from 2S and the (2n-1)/2^n from S are left with the sequence (2/2 +2/4 +2/8....)

2S – S = 1 + 2/2 + 2/4 + 2/8 + …2/2^n - (2n-1)/2^n

S = 1 + 2*( 1/2 + 1/4 + 1/8 …) - (2n-1)/2^n

now i know that 2*( 1/2 + 1/4 + 1/8 …) =2*1=2
so

S=1+2-(2n-1)/2^n
=3-(2n-1)/2^n

only that the answer is wrong, and the correct one is

3-(2n+3)/2^n

can you see where i have gone wrong??