Question

The sum to infinity of a geometric series is 32.The sum of the first four terms is 30 and

all the terms are positive.Find the difference between the sum to infinity and the sum
of the first eight terms

Answer 1

The terms in the series are

`\displaystyle \sum_(n=1)^\infty ar^(n-1) = a + ar + ar^2 + ar^3 + \cdots`

The series converges, so we
`|r|<1`. All the terms in this particular series are known to be positive, so
`0</p><p>Consider the [tex]N`-th partial sum of the series,

`S_N = \displaystyle \sum_(n=1)^N ar^(n-1) = a + ar + ar^2 + \cdots + ar^(N-1)`

Multiply both sides by
`r` :

`r S_N = ar + ar^2 + ar^3 + \cdots + ar^N`

Subtracting this from
`S_N` eliminates all the
`r^n` terms between
`n=1` and
`n=N-1`, leaving us with

`(1 - r) S_N = a - ar^N \implies S_N = (a(1-r^N))/(1-r)`

As
`N\to\infty`, the
`r^N` term will converge to zero, and the value of the infinite series is

`\displaystyle \sum_(n=1)^\infty ar^(n-1) = \lim_(n\to\infty) S_N = S = \frac a{1-r}`

Now, we're given

`S = \frac a{1-r} = 32 ~~~~ \text{ and } ~~~~ S_4 = (a(1-r^4))/(1-r) = 30`

By substitution, we have

`S_4 = 32(1-r^4) = 30 \implies 1-r^4 = (15)/(16) \implies r^4 = \frac1{16} \implies r = \frac12`

so that

`S = \frac a{1 - \frac12} = 32 \implies a = 16`

Then the sum of the first 8 terms is

`S_8 = (a(1-r^8))/(1-r) = \frac{16\left(1 - \frac1{2^8}\right)}{1-\frac12} = \frac{255}8`

and the difference between this and the infinite sum is

`S - S_8 = 32 - \frac{255}8 = \frac{256-255}8 = \boxed{\frac18}`