§8.3 Geometric sums (continued)

How geometric series naturally connect to Taylor polynomials

Summary

Differentiating $\frac{1}{1-x} = 1+x+x^2+\cdots$ term by term gives $\frac{1}{(1-x)^2} = 1+2x+3x^2+\cdots$. Differentiating again would produce

For $f(x) = \frac{1}{1-x}$, every derivative at zero equals $1$, so the Taylor series is $1+x+x^2+x^3+\cdots$. The degree $n$ Taylor polynomial is

The Taylor polynomials $T_n(x) = 1+x+\cdots+x^n$ of $\frac{1}{1-x}$ relate to the geometric partial sums by

The equality $\frac{1}{1-x} = 1+x+x^2+\cdots$ holds

For the borderline ratios $r = -1$ and $r = 2$, the partial sums of $\sum ar^k$ satisfy

A ball dropped from $16$ meters rebounds to half the height each bounce. The total vertical distance $16+2\cdot8+2\cdot4+2\cdot2+\cdots$ is a geometric series. Report the total distance in meters.

A worker is paid $1$ cent on day one, doubling daily. The ten-day total in dollars is $0.01\left(2^{10}-1\right)$. Report this value.

Continuing the doubling pay schedule, the thirty-day total in dollars is $0.01\left(2^{30}-1\right)$. Report this value.

The series $1+1+1+\cdots$ has ratio $r = 1$, so its partial sums $S_n = n$ grow without bound and the series diverges.