Vtyjkyuk

I just need the formula for the sum of geometric series when each element in the series has the value $1/2^j+1$, where $j = 0, 1, 2, ldots, n$. Please help.

Someone told me it is:

$$S = 2 - frac12^n$$

I am not sure if its right because he has given me no proof and I couldn't prove it when I calculate it manually. Say for example:

$$S = 1/2 + 1/4 + 1/8 = .875$$

But when using the formula given above, with $n=3$ (since there are $3$ elements):

$$S = 2 - 1/8 = 1.875$$

The answers are not the same. Please enlighten me with this issue.

Consider the $n$th partial sum
$$S_n = sum_j = 0^n r^j = 1 + r + r^2 + cdots + r^n$$
of the geometric series
$$sum_j = 0^infty r^j$$
with common ratio $r$.

If we multiply $S_n$ by $1 - r$, we obtain
beginalign*
(1 - r)S_n & = (1 - r)(1 + r + r^2 + cdots + r^n)\
& = 1 + r + r^2 + cdots + r^n - (r + r^2 + r^3 + cdots + r^n + 1)\
& = 1 - r^n + 1
endalign*
If $r neq 1$, we may divide by $1 - r$ to obtain
$$S_n = frac1 - r^n + 11 - r$$
In particular, if $r = 1/2$, we obtain
beginalign*
S_n & = sum_r = 0^n left(frac12right)^j\
& = frac1 - left(frac12right)^n + 11 - frac12\
& = frac1 - left(frac12right)^n + 1frac12\
& = 2left[1 - left(frac12right)^n + 1right]\
& = 2left(1 - frac12^n + 1right)\
& = 2 - frac12^n
endalign*
which is the formula you were given.

However, you want
beginalign*
sum_j = 0^n + 1 frac12^j + 1 & = frac12 sum_j = 0^n frac12^j\
& = frac12 sum_j = 0^n left(frac12right)^n\
& = frac12left[2 - frac12^nright]\
& = 1 - frac12^n + 1
endalign*
As a check, observe that when $n = 2$
$$sum_j = 0^2 frac12^j + 1 = frac12 + frac14 + frac18 = frac78 = 0.875$$
and
$$1 - frac12^2 + 1 = 1 - frac12^3 = 1 - frac18 = frac78 = 0.875$$
In your calculation, you used $n = 3$ because you did not take into account the fact that if the index starts with $0$, the third term is $n = 2$.

You are computing the sum$$S_n=frac12+frac14+frac18+cdots+frac12^n+1.$$Well,$$2S_n=1+frac12+frac14+frac18+cdots+frac12^n=S_n+1-frac12^n+1$$and therefore$$S_n=1-frac12^n+1.$$