Given the sequence $(x_n)$, what is $\sum_{k=0}^\infty 1/x_k$?

Given the sequence $(x_n)$, what is $\sum_{k=0}^\infty 1/x_k$?

Question

The sequence $(x_n)$, where $n \geq 0$, is defined by $x_0 =1$ and

$$x_n = \sum_{k=0}^{n-1}x_k \quad \text{for} \quad n \geq 1.$$

The sum

$$\sum_{k=0}^\infty \dfrac{1}{x_k}$$

equals

1. $1$, (b) $\dfrac{6}{5}$, (c) $\dfrac{8}{5}$, (d) $3$, (e) $\dfrac{27}{5}$.

Suggestion

Can we write the sequence $x_0, x_1, x_2,...$ out? What kind of a sequence is this (almost)?

So what kind of a sequence is $\dfrac{1}{x_n}$ (almost)?

How do we sum this kind of sequence?

Solution

Can we write the sequence $(x_n)$ out?

Using the rule, we have:

$x_0 = 1$,

$x_1 =x_0= 1$,

$x_2=x_0+x_1=1+1=2$,

$x_3 =x_0+x_1+x_2= 1+1+2 = 4$,

$x_4=x_0+x_1+x_2+x_3 = 1+1+2+4 = 8, ...$

So $x_n = 2^{n-1}$ for $n \geq 1$, and apart from the first term, we have a geometric sequence.

We also have that $\dfrac{1}{x_n} = \dfrac{1}{2^{n-1}}$, and so the sequence $\left(\dfrac{1}{x_n}\right)$ is also geometric except for the first term.

We can remember now that the sum to infinity of a geometric progression with first term $a$ and common ratio $r$ is $\dfrac{a}{1-r}$.

So $\sum_{k=0}^\infty \dfrac{1}{x_k} = 1 + 1 + \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...$

$$= 1 + \dfrac{1}{1-\dfrac{1}{2}} = 3,$$

and so that answer is (d).