The terms in the series are
[tex]\displaystyle \sum_{n=1}^\infty ar^{n-1} = a + ar + ar^2 + ar^3 + \cdots[/tex]
The series converges, so we [tex]|r|<1[/tex]. All the terms in this particular series are known to be positive, so [tex]0
Consider the [tex]N[/tex]-th partial sum of the series,
[tex]S_N = \displaystyle \sum_{n=1}^N ar^{n-1} = a + ar + ar^2 + \cdots + ar^{N-1}[/tex]
Multiply both sides by [tex]r[/tex] :
[tex]r S_N = ar + ar^2 + ar^3 + \cdots + ar^N[/tex]
Subtracting this from [tex]S_N[/tex] eliminates all the [tex]r^n[/tex] terms between [tex]n=1[/tex] and [tex]n=N-1[/tex], leaving us with
[tex](1 - r) S_N = a - ar^N \implies S_N = \dfrac{a(1-r^N)}{1-r}[/tex]
As [tex]N\to\infty[/tex], the [tex]r^N[/tex] term will converge to zero, and the value of the infinite series is
[tex]\displaystyle \sum_{n=1}^\infty ar^{n-1} = \lim_{n\to\infty} S_N = S = \frac a{1-r}[/tex]
Now, we're given
[tex]S = \dfrac a{1-r} = 32 ~~~~ \text{ and } ~~~~ S_4 = \dfrac{a(1-r^4)}{1-r} = 30[/tex]
By substitution, we have
[tex]S_4 = 32(1-r^4) = 30 \implies 1-r^4 = \dfrac{15}{16} \implies r^4 = \dfrac1{16} \implies r = \dfrac12[/tex]
so that
[tex]S = \dfrac a{1 - \frac12} = 32 \implies a = 16[/tex]
Then the sum of the first 8 terms is
[tex]S_8 = \dfrac{a(1-r^8)}{1-r} = \dfrac{16\left(1 - \frac1{2^8}\right)}{1-\frac12} = \dfrac{255}8[/tex]
and the difference between this and the infinite sum is
[tex]S - S_8 = 32 - \dfrac{255}8 = \dfrac{256-255}8 = \boxed{\dfrac18}[/tex]
