Divide by Prime

Calculus Level 3

A sequence of real numbers $\{ a _n \}$ is defined using the following recurrence relation

$a _{ n + 1} = \frac{ a_{n } } { p_{ n } }$

for all $n \geq 1$ . If $p_{n}$ denotes the $n^\text{th}$ prime number, what is $\displaystyle \lim _{ n \to \infty } a_{n}$ ?

It depends on the initial value,

a_1

1 The sequence does not converge -1 0

1 solution

Pranshu Gaba
Jan 19, 2016

In the problem, we are given a recurrence relation for $a _{ n}$ . We will now find an explicit expression for $a _ {n }$ which will help us understand the sequence better.

$a _{n} = \frac{ a_{n - 1 } }{ p _{n - 1 } } = \frac { a _{n - 2} } { p _{ n- 1 } \times p _{n - 2 } } = \frac { a _{n - 3} } { p _{ n- 1 } \times p _{n - 2 } \times p _{n - 3 } } = \cdots = \frac{ a _ {1} } { p_{1} \times p_{2} \times p_{3} \times \cdots \times p_{n - 1 }}$

Irrespective of what $a_ 1$ is, we see that as $n$ becomes bigger, more and more large numbers are multiplied in denominator. However, the numerator remains constant so the fraction becomes smaller and approaches $\boxed{ 0 }$ $_\square$

Moderator note:

Simple standard approach.

Divide by Prime

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