Sequences and Series – Proof of Monotone and Bounded Sequences

Question

Say $s_1=1$ and $s_{n+1}=\frac{1}{5}(s_n+7)$ for $n\geq 1$. Prove that the sequence is monotone and bounded, then find the limit.

Answer 1

We have $$ s_{n+1}-s_n=\frac15(s_n-s_{n-1}) \quad \forall n \ge 2, $$ and so we have $$\tag{*} s_{n+1}-s_n=\frac{s_2-s_1}{5^{n-1}}=\frac{3}{5^n} \quad \forall n \ge 1. $$ Thus $s_n<s_{n+1}$ for every $n \ge 1$, i.e. $(s_n)$ is increasing.

Thanks to (*) we have for every $n \ge 1$: \begin{eqnarray} s_n&=&1+\sum_{i=1}^{n-1}(s_{i+1}-s_i)=1+\sum_{i=1}^{n-1}\frac{3}{5^i}=-2+3\sum_{i=0}^{n-1}\frac{1}{5^i}=-2+3\cdot\frac{1-\frac{1}{5^n}}{1-\frac15}\\ &=&-2+\frac{15}{4}\left(1-\frac{1}{5^n}\right)=\frac74-\frac34\cdot\frac{1}{5^{n-1}}<\frac74. \end{eqnarray} We deduce that $(s_n)$ is a convergent sequence and $$ \lim_ns_n=\lim_n\left(\frac74-\frac34\cdot\frac{1}{5^{n-1}}\right)=\frac74. $$