Arithmetico-geometric progressions

Some series are nei­ther arith­metic nor geo­met­ric, but are built from one of each. In $1 + 2 \cdot 2 + 3 \cdot 4 + 4 \cdot 8 + \cdots$ the first fac­tors $1, 2, 3, 4$ form an A.P. and the sec­ond fac­tors $1, 2, 4, 8$ form a G.P. Such a series is called arith­metico-geo­met­ric. The for­mu­las of the ear­lier lessons do not apply directly, but the trick used for the sum of a G.P., mul­ti­ply­ing by $r$ and sub­tract­ing, still works.

Def­i­n­i­tion#

A sequence is called an arith­metico-geo­met­ric pro­gres­sion (A.G.P.) if each term is the prod­uct of the cor­re­spond­ing terms of an A.P. and a G.P.

If the A.P. is $a, a + d, a + 2d, \ldots$ and the G.P. is $1, r, r^2, \ldots$, the A.G.P. is

$$a, \quad (a + d)r, \quad (a + 2d)r^2, \quad (a + 3d)r^3, \quad \ldots$$

The gen­eral term#

The $n$th term is the $n$th term of the A.P. times the $n$th term of the G.P.:

$$t_n = \left[a + (n - 1)d\right] r^{n-1}$$

Exam­ple 1#

Find the gen­eral term and the 5th term of $2 \cdot 1 + 5 \cdot 3 + 8 \cdot 9 + \cdots$

The A.P. is $2, 5, 8, \ldots$ with $n$th term $3n - 1$. The G.P. is $1, 3, 9, \ldots$ with $n$th term $3^{n-1}$. So $t_n = (3n - 1) \cdot 3^{n-1}$, and $t_5 = 14 \times 81 = 1134$.

The sum of n terms: the method#

Here is the method on a par­tic­u­lar series, then the gen­eral result.

Exam­ple 2#

Find the sum of $n$ terms of $1 + 2 \cdot 2 + 3 \cdot 2^2 + 4 \cdot 2^3 + \cdots$

Write the sum, then the sum mul­ti­plied by the com­mon ratio 2, shifted one place to the right:

$$\begin{aligned} S &= 1 + 2 \cdot 2 + 3 \cdot 2^2 + \cdots + n \cdot 2^{n-1} \\ 2S &= \phantom{1 + {}} 1 \cdot 2 + 2 \cdot 2^2 + \cdots + (n - 1) \cdot 2^{n-1} + n \cdot 2^n \end{aligned}$$

Sub­tract the sec­ond line from the first. Terms in the same col­umn dif­fer by exactly one power of 2:

$$S - 2S = 1 + 2 + 2^2 + \cdots + 2^{n-1} - n \cdot 2^n$$

The mid­dle part is now a plain G.P. with $n$ terms, whose sum is $2^n - 1$. So

$$-S = 2^n - 1 - n \cdot 2^n \implies S = (n - 1) 2^n + 1$$

Check with $n = 4$: $1 + 4 + 12 + 32 = 49$, and $(4 - 1) \times 16 + 1 = 49$.

The sum of n terms: the gen­eral result#

The same steps for $S_n = a + (a + d)r + \cdots + \left[a + (n - 1)d\right]r^{n-1}$ give

$$S_n - rS_n = a + d(r + r^2 + \cdots + r^{n-1}) - \left[a + (n - 1)d\right]r^n$$

and so, for $r \neq 1$,

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

There is no need to mem­o­rise this. It is quicker and safer to repeat the method each time: write $S$, write $rS$ shifted one place, sub­tract, and sum the G.P. that appears.

The sum to infin­ity#

If $-1 < r < 1$, then $r^n$ shrinks to 0 as $n$ grows, and so does $n r^n$ (the pow­ers shrink much faster than $n$ grows). The last term of the for­mula dis­ap­pears, and $r^{n-1}$ dis­ap­pears from the mid­dle term, leav­ing

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

Exam­ple 3#

Find the sum to infin­ity of $\displaystyle 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \cdots$

By the method: let $\displaystyle S = 1 + 2 \cdot \frac{1}{3} + 3 \cdot \frac{1}{3^2} + \cdots$. Then

$$\displaystyle \frac{1}{3}S = \frac{1}{3} + 2 \cdot \frac{1}{3^2} + 3 \cdot \frac{1}{3^3} + \cdots$$

Sub­tract­ing, $\displaystyle S - \frac{1}{3}S = 1 + \frac{1}{3} + \frac{1}{9} + \cdots = \frac{1}{1 - 1/3} = \frac{3}{2}$. So $\displaystyle \frac{2}{3}S = \frac{3}{2}$ and $\displaystyle S = \frac{9}{4}$.

By the for­mula, with $a = 1$, $d = 1$, $\displaystyle r = \frac{1}{3}$: $\displaystyle \frac{1}{2/3} + \frac{1/3}{4/9} = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}$. The two agree.

Exam­ple 4#

Find the sum to infin­ity of $\displaystyle 1 + \frac{4}{5} + \frac{7}{25} + \frac{10}{125} + \cdots$

The numer­a­tors $1, 4, 7, 10$ form an A.P. with $a = 1$, $d = 3$; the denom­i­na­tors give the G.P. with $\displaystyle r = \frac{1}{5}$.

$$\displaystyle S_\infty = \frac{1}{1 - \frac{1}{5}} + \frac{3 \cdot \frac{1}{5}}{\left(1 - \frac{1}{5}\right)^2} = \frac{5}{4} + \frac{3/5}{16/25} = \frac{5}{4} + \frac{15}{16} = \frac{35}{16}$$

Exam­ple 5#

Find the sum to infin­ity of $\displaystyle 3 + \frac{5}{2} + \frac{7}{4} + \frac{9}{8} + \cdots$

Here $a = 3$, $d = 2$, $\displaystyle r = \frac{1}{2}$:

$$\displaystyle S_\infty = \frac{3}{1/2} + \frac{2 \cdot \frac{1}{2}}{1/4} = 6 + 4 = 10$$

Exam­ple 6#

Show that $\displaystyle 1 + 2x + 3x^2 + 4x^3 + \cdots = \frac{1}{(1 - x)^2}$ when $-1 < x < 1$.

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

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

Prac­tice#

1. Write the gen­eral term and the 6th term of $1 \cdot 2 + 3 \cdot 4 + 5 \cdot 8 + \cdots$
2. Write the first four terms of the A.G.P. with $a = 2$, $d = 3$ and $\displaystyle r = \frac{1}{2}$.
3. Find the sum to infin­ity of $\displaystyle 1 + \frac{2}{2} + \frac{3}{4} + \frac{4}{8} + \cdots$
4. Find the sum to infin­ity of $\displaystyle 2 + \frac{5}{3} + \frac{8}{9} + \frac{11}{27} + \cdots$
5. Find the sum to infin­ity of $\displaystyle 1 + \frac{3}{4} + \frac{5}{16} + \frac{7}{64} + \cdots$
6. Find the sum of $n$ terms of $1 + 2 \cdot 3 + 3 \cdot 3^2 + \cdots$, and check it with $n = 3$.
7. Use Exam­ple 6 to find $\displaystyle 1 + \frac{2}{4} + \frac{3}{16} + \frac{4}{64} + \cdots$
8. The sum to infin­ity of $\displaystyle 3 + \frac{3 + d}{4} + \frac{3 + 2d}{16} + \cdots$ is $\displaystyle \frac{44}{9}$. Find $d$.

Answers#