How to prove that if $\lim_{n\to\infty} a_n \prod_{k=1}^n b_k = l>0$ then, for any $0<c_k<b_k$, $\lim_{n\to\infty} a_n \prod_{k=1}^n (b_k-c_k) < l$?

Question

How to prove that if $$\lim_{n\to\infty} a_n \prod_{k=1}^n b_k = l>0$$ then for any $0<c_k<b_k$ $$ \ \ \ \ \ \ \ \ \ \ \lim_{n\to\infty} a_n \prod_{k=1}^n (b_k-c_k) < l \ \ \ \ \ ?$$

My professor mentioned this to me today but didn't prove it. I agree with him that it is intuitive, but since we're dealing with a limit we "could" (*) have $c_k$ converge to $0$ so fast that the statement doesn't hold, so I'm looking for a proof, which I really cannot think of, I guess I'd need more familiarity with infinite products.

(*) In fact the statement is supported by numerical evidence even in that sense.

Answer 1

Hint: Put $\displaystyle u_n=a_n\prod_{k=1}^n b_k$, $\displaystyle v_n=a_n\prod_{k=1}^n (b_k-c_k)$ You have for $n\geq 2$:

$$\frac{v_n}{u_n}=\prod_{k=1}^n (1-\frac{c_k}{b_k})=(1-\frac{c_1}{b_1})\prod_{k=2}^n (1-\frac{c_k}{b_k})\leq (1-\frac{c_1}{b_1})$$