Prove $(s_n)$ converges, where $s_{n+1} \lt r(s_n)$.

Question

Suppose $(s_n)$ is a sequence such that $s_{n+1} < r s_n$ where $r$ is a constant with $0<r<1$. Prove that $s_n$ converges. All values are positive.

I thought about maybe solving this through the Cauchy criterion and setting $s_{n+1} - r s_n < 0$ but not sure how to put an absolute value on it and setting it to be less than epsilon.

Answer 1

Lets consider the series $S = \sum_{n=1}^\infty s_n$, since we have that $$\dfrac{s_{n+1}}{s_n} < r < 1$$ we know $S$ converges (ratio test), hence what can we conclude about $\lim_{n \to \infty} s_n$?

Answer 2

By induction, $0<s_n<r^ns_0\to 0$

Answer 3

By assumption $$ s_{n+1} < r s_{n} $$ for all $n$ with $s_n >0, r \in (0,1)$. Thus, using this iteratively, for each $n \in \mathbb{N}$ we get $$ s_{n+1} < r s_{n} < r^2 s_{n-1} < r^3 s_{n-2} < ... <r^{n+1} s_0. $$ Since $r^n \rightarrow 0$ as $n \rightarrow \infty$ we see that also $s_n \rightarrow 0$ as $n \rightarrow \infty$, which proves the claim.