Question

What is the sum of the first `n` terms of the series `1/2+3/4+5/8+...` ?

Answer 1

`S_n=3-(2n+3)/2^n`

Explanation:

Let `S_n=1/2+3/4+5/8+.................."n terms"`

What is `n^(th)` term of the series? Well numerators are in arithmatic sequence and numerator of `n^(th)` term is `2n-1` and denominator is in geometric series with `n^(th)` term being `2^n`. Hence `n^(th)` term of given series is `(2n-1)/2^n` and series is

`S_n=1/2+3/4+5/8+7/16+..................+(2n-1)/2^n` and then

`1/2S_n=1/4+3/8+5/16+..................+(2n-1)/2^(n+1)`

Subtracting latter from former, we get

`S_n-1/2S_n=1/2+2/4+2/8+....+2/2^n-(2n-1)/2^(n+1)`

or `1/2S_n=1/2+2/4[1+1/2+....+1/2^(n-2)]-(2n-1)/2^(n+1)`

= `1/2+1/2(1-1/2^(n-1))/(1-1/2)-(2n-1)/2^(n+1)`

= `1/2+1-1/2^(n-1)-(2n-1)/2^(n+1)`

= `3/2-(4+2n-1)/2^(n+1)`

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

and `S_n=3-(2n+3)/2^n`