Question

alpha writes the infinite arithmetic sequence \[10, 8, 6, 4, 2, 0 ,\ldots.\]beta writes the infinite geometric sequence \[9, 6, 4, \frac{8}{3}, \frac{16}{9}, \ldots.\]gamma makes a sequence whose $n^{\text{th}}$ term is the product of the $n^{\text{th}}$ term of alpha's sequence and the $n^{\text{th}}$ term of beta's sequence: \[10\cdot 9 \quad,\quad 8\cdot 6\quad ,\quad 6\cdot 4\quad,\quad 4\cdot \frac83\quad,\quad 2\cdot \frac{16}{9}\quad,\quad \ldots.\]what is the sum of gamma's entire sequence?

Answer 1

Consecutive terms in the α sequence have a common difference of

8 - 10 = 6 - 8 = 4 - 6 = … = -2

so they are given recursively by

$\begin{cases} \alpha_1 = 10 \\ \alpha_n = \alpha_{n-1} - 2 & \text{for } n\ge2 \end{cases}$

By substitution, we have

$\alpha_n = \alpha_{n-1} - 2 \\\\ \alpha_n = \alpha_{n-2} - 2\cdot2 \\\\ \alpha_n = \alpha_{n-3} - 3\cdot2 \\\\ \vdots \\\\ \alpha_n = \alpha_{n-(n-1)} - (n-1)\cdot 2 = \alpha_1 - 2(n-1) = 12-2n$

Consecutive terms in the β sequence have a common ratio of

6/9 = 4/6 = (8/3)/4 = … = 2/3

so the recurrence for these terms is

$\begin{cases} \beta_1 = 9 \\\\ \beta_n = \frac23 \beta_{n-1} & \text{for } n\ge2 \end{cases}$

We can solve for $\beta_n$ similarly:

$\beta_n = \dfrac23 \beta_{n-1} \\\\ \beta_n = \left(\dfrac23\right)^2 \beta_{n-2} \\\\ \beta_n = \left(\dfrac23\right)^3 \beta_{n-3} \\\\ \vdots \\\\ \beta_n = \left(\dfrac23\right)^{n-1} \beta_{n-(n-1)} = \left(\dfrac23\right)^{n-1} \beta_1 = \dfrac{2^{n-1}}{3^{n-1}} \cdot 3^2 = \dfrac{2^{n-1}}{3^{n-3}}$

The γ sequence has $n$-th term

$\gamma_n = \alpha_n \beta_n = (12-2n) \dfrac{2^{n-1}}{3^{n-3}} = (108-18n) \left(\dfrac23\right)^{n-1}$

and we want to compute

$\displaystyle \sum_{n=1}^\infty \gamma_n = 108 \sum_{n=1}^\infty \left(\frac23\right)^{n-1} - 18 \sum_{n=1}^\infty n \left(\frac23\right)^{n-1}$

Recall the sum of an infinite geometric series with common ratio $|r|$ converges to

$\displaystyle \sum_{n=1}^\infty r^{n-1} = \frac1{1-r}$

so that

$\displaystyle \sum_{n=1}^\infty \left(\frac23\right)^{n-1} = \frac1{1-\frac23} = 3$

For the remaining sum, we can use the method shown in question [24494877] to compute

$\displaystyle \sum_{n=1}^\infty nr^{n-1} = \frac1{(1-r)^2}$

which gives

$\displaystyle \sum_{n=1}^\infty n \left(\frac23\right)^{n-1} = \frac1{\left(1-\frac23\right)^2} = 9$

Then the infinite sum of the terms of γ converges to

$\displaystyle \sum_{n=1}^\infty \gamma_n = 108 \cdot 3 - 18 \cdot 9 = \boxed{162}$