Exercise 4.6 — Question 1

Key Summation Formulas

The following standard results are used throughout Exercise 4.6:

${\sum }_{k=1}^{n}k=\frac{n(n+1)}{2}$

${\sum }_{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$

${\sum }_{k=1}^n{k}^3={\left[\frac{n(n+1)}{2}\right]}^2$

Arithmetic-Geometric Sequences

An arithmetic-geometric sequence is formed by multiplying corresponding terms of an arithmetic sequence and a geometric sequence.

If the arithmetic sequence has first term $a$ and common difference $d$, and the geometric sequence has first term $1$ and common ratio $r$, then the general term is:

$T_n=[a+(n-1)d]r^{n-1}$

Sum to $n$ Terms

The sum $S_n$ of an arithmetic-geometric series is found using the multiply-and-subtract method:

$S_n=\frac{a-[a+(n-1)d]r^n}{1-r}+\frac{dr(1-r^{n-1})}{(1-r{)}^2},r\ne 1$

Sum to Infinity

When $|r|<1$, the series converges and:

$S_{\infty }=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$

Sigma Notation

Sigma notation is used to write series compactly. For example:

${\sum }_{k=1}^{n}k{r}^{k-1}=1+2r+3r^2+\cdots +n{r}^{n-1}$

Worked Example

Find the sum of the series $1+2r+3r^2+4r^3+\cdots$ to infinity, given $|r|<1$.

Solution:

This is an arithmetic-geometric series with $a=1$, $d=1$, and common ratio $r$.

Using the formula for $S_{\infty }$:

$S_{\infty }=\frac{1}{1-r}+\frac{r}{(1-r{)}^2}=\frac{1-r+r}{(1-r{)}^2}=\frac{1}{(1-r{)}^2}$