Where have i gone wrong with this lim

[Dell]

where am i going wrong with this lim?
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


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??

 

[quantumdude]

I don't understand why the sum isn't just 3. You clearly have:

$$S=1+2\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^n$$

The sum of the geometric series above is known to be exactly 1, so S=1+2(1)=3, no?

 

[Dell]

correct, but as i have shown before, when i subtract S from 2S i subtract the 'n+1' number of S from the 'n' number of 2S, leaving me with the 1st from 2S and the last from S, with the geometric series between, if you look at the answer '3-(2n+3)/2^n' it is correct for ANY n,
n=1, S=1/2
n=2 S=7/4...

how do i get to this?