A sequence of positive numbers strictly decreasing converges to 0?

Question

Suppose ${ \{s_n \}}_{n=1}^\infty $

be a sequence of positive numbers and $0<x<1$ . If $s_{n+1}<xs_n (n\in I)$ then $$\lim_{n\rightarrow \infty} s_n =0 $$

My attempt: since $x<1$ it follows that $s_{n+1}<s_n $. Thus $s_n$ is a decreasing sequence and also bounded below by 0. Therefore it is convergent.

But to say that this sequence exactly converge to 0 i tried contradiction method like:

If there exists no N such that $$|s_n|< \epsilon , (n\ge N)$$

Thus for any N $$|s_n| > \epsilon, (n\ge N) $$

$$\implies s_n > \epsilon, (n\ge N)$$

But then w.k.t. $$s_{N+1}<s_N$$

Is this gives a contradiction or did i do something wrong? Can anyone explain!!

Answer 1

Let $\varepsilon>0$ and $0<x<1$, $x\in \mathbb{R}$. From the given sequence and the condition $s_{n+1}<xs_n$, as @kieransquared mentioned in the comments, we can see that $s_{n+1}<x^ns_1$ or similarly $s_{n}<x^{n-1}s_1$. This means that there exists $N$ such that $x^Ns_1<\varepsilon$, since all members of the given sequence are positive real numbers, as such $s_1$ is also a positive real number. Then $\forall n > N$ we have:$$|s_{n}-0|=s_n<x^{n-1}s_1<x^Ns_1<\varepsilon\implies N>\log_{x}\frac{\varepsilon}{s_1}$$

Since $\varepsilon>0$ is arbitrary we come to the conclusion that $\forall \varepsilon >0 \exists N=\lfloor\log_{x}\frac{\varepsilon}{s_1}\rfloor+1, \forall n >N $ $|s_n-0|<\varepsilon$. This means $\displaystyle \lim_{n\to \infty}{s_n}=0$.

Answer 2

The squeeze theorem for sequences states that:

Let $\displaystyle (a_{n}),(c_{n}) $ be two sequences converging to $\ell$, and $(b_{n})$ a sequence. If $\displaystyle \forall n\geq N,N\in \mathbb {N} $ we have $\displaystyle a_{n}\leq b_{n}\leq c_{n}$, then $(b_{n})$ also converges to $\ell$.

Since $0<x<1,\ \displaystyle\lim_{n\to\infty} x^n = 0.\ $ Next, we have that $0\leq s_1,\ $ $0\leq s_2 < x s_1,\ 0\leq s_3 < x s^2 < x^2 s_1,\ \ldots.$

For all $n\in\mathbb{N},$ we have: $0\leq s_n < x^{n-1} s_1.$

By the squeeze theorem above with $a_n = 0,\ b_n = s_n $ and $\ c_n = x^{n-1} s_1\ ,\ $ the result follows.