Irrational sum

Algebra Level 4

Let $a_{1}, a_{2}, ..., a_{2019}$ and $b_{0}, b_{1}, ..., b_{2019}$ be real numbers different from 0, which satisfy $b_{n} = \left(1 -\dfrac{1}{a_{n}} \right)b_{n-1}$ for $n = 1,...,2019.$ Assume that the sum

$S = \frac{1}{a_{1}b_{1}} + \frac{1}{a_{2}b_{2}} + ... + \frac{1}{a_{2019}b_{2019}}$

is an irrational number. Can both $b_{0}$ and $b_{2019}$ be rational numbers?

Cannot be determined Yes No

1 solution

Magnus Seidenfaden
Mar 7, 2019

$b_{n} = (1 -$ $\frac{1}{a_{n}}$ $) b_{n-1}$ We divide through with $b_{n}b_{n-1}$ :

$\frac{1}{b_{n-1}}$ = $( 1 -$ $\frac{1}{a_{n}}$ $)$ $\frac{1}{b_{n}}$

$\frac{1}{a_nb_n}$ = $\frac{1}{b_n}$ $-$ $\frac{1}{b_{n-1}}$

Thus the entire sum is just a telescoping series resulting in $S =$ $\frac{1}{b_{2019}}$ $-$ $\frac{1}{b_{0}}$

If S has to be irrational at least one of $b_{2019}$ and $b_0$ has to be irrational.

Irrational sum

1 solution

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