Question

Find the infinite sum, if it exists: `–2+1/2-1/8+1/32-...`?

Answer 1

`"S"_oo=-8/5`

Explanation:

For a geometric sequence, `a, ar, ar^2...` where `a` is the first term and `r` is the common ratio between two consecutive terms, the sum to infinity, `"S"_oo,` is given in this equation, `"S"_oo=a/(1-r)`.

The given series is geometric because the common ratio, `r`, is `-1/4`.

Therefore, `"S"_oo=(-2)/(1--1/4)=-8/5`

Answer 2

The infinite sum exists; `S=–8/5`.

Explanation:

The sequence converges, since each term is `-1/4` times the previous term, and `abs(-1/4) < 1`. Thus the infinite sum exists. Let us call this sum `S`. Then:

`color(white)(–1/2)S=-2+1/2-1/8+...`

`color(white)(–1/2S)=-2(1-1/4+1/16-1/64+...)`

`color(blue)(–1/2 S="    "1-1/4+1/16-1/64+...)`

If we multiply both sides by `-1/4`, we get

`(–1/4)(–1/2 S)=(–1/4)(1-1/4+1/16-1/64+...)`

`color(green)("                     "1/8 S = -1/4 + 1/16 - 1/64 + ...)`

Notice how the expansion for `color(blue)(-1/2 S)` includes the expansion for `color(green)(1/8 S)`. If we subtract `1/8 S` from `-1/2 S` we get:

`"      "color(blue)(–1/2 S) - color(green)(1/8 S)=color(blue)("   "1-1/4+1/16-1/64+...)`
`color(white)(–1/2 S - 1/8 S=)color(green)("     "-(-1/4+1/16-1/64+...))`

`"                  "–5/8 S=1`

Solving for `S` gives:

`S=-8/5`