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

2 Answers
Apr 16, 2017

"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

Apr 16, 2017

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