Exercise 4.5 — Question 4

Summation Formulas (∑n, ∑n², ∑n³)

The following standard results are used to evaluate series involving natural numbers:

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

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

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

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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:

1. Write $S_n=a+(a+d)r+(a+2d)r^2+\cdots +[a+(n-1)d]r^{n-1}$
2. Multiply both sides by $r$: $r{S}_n=ar+(a+d)r^2+\cdots +[a+(n-1)d]r^n$
3. Subtract: $(1-r)S_n=a+d(r+r^2+\cdots +r^{n-1})-[a+(n-1)d]r^n$
4. Use the geometric series formula to simplify.

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

Sum to Infinity

When $|r|<1$, as $n\to \infty$, $r^n\to 0$, so:

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

Sigma Notation

The arithmetic-geometric series can be written as:

${\sum }_{k=1}^n[a+(k-1)d]r^{k-1}$

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Worked Example

Find the sum to $n$ terms and sum to infinity of: $1+2r+3r^2+4r^3+\cdots$ where $|r|<1$.

Here $a=1$, $d=1$:

$S_n=\frac{1-(n+1)r^n}{1-r}+\frac{r(1-r^{n-1})}{(1-r{)}^2}$

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