Prove product of sequence less than other for all n using induction

53 Views Asked by At

Given two sequences $a_n$ and $b_n$, with the former always smaller than the latter, I want to prove that $\prod_{i=1}^{n}a_i \le \prod_{i=1}^{n}b_i$.

I know I'm going to use induction on n, and have verified the base case, with n=0. However, I am having trouble with how to proceed. How can I get started with showing $\prod_{i=1}^{k+1}a_i \le \prod_{i=1}^{k+1}b_i$ from $\prod_{i=1}^{k}a_i \le \prod_{i=1}^{k}b_i$?

1

There are 1 best solutions below

2
On

Assuming ${a_n}, {b_n}$ are non-negative, we have.

$\prod_{i=1}^{k}a_i \le \prod_{i=1}^{k}b_i \implies \prod_{i=1}^{k+1}a_{i} = \prod_{i=1}^{k}a_{i} * a_{k+1} \le \prod_{i=1}^{k}a_i * b_{k+1} \le \prod_{i=1}^{k}b_i * b_{k+1} = \prod_{i=1}^{k+1}b_i$