Successions and series, exercise. It is true that
if $\;a_n>0\;$ and $\;\lim\limits_{n\to\infty}\dfrac{a_{n + 1}}{ a_n}<1\;,\;$ then $\;\lim\limits_{n\to\infty}a_n=0\;?$
Is that true or false?
 A: Yes, it is true.
Since $\;\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}<1\;,\;$ there exist $\;h\in\left]0,1\right[\;$ and $\;\nu\in\mathbb{N}\;$ such that
$\dfrac{a_{n+1}}{a_{n}}<h\quad$ for all $\;n>\nu\;.\quad\color{blue}{(*)}$
Since $\;a_n>0\;$ for all $\;n\in\mathbb{N}\;,\;$ from $(*)$ we get that
$0<a_{n+1}<h^{n-\nu}a_{\nu+1}\;$ for all $\;n>\nu\;.\quad\color{blue}{(**)}$
Moreover, $\;\lim\limits_{n\to\infty}h^{n-\nu}a_{\nu+1}=0\;$ and, by applying the squeeze theorem to the inequalities $(**)$, we get that
$\lim\limits_{n\to\infty}a_{n+1}=0\;,\;$ hence
$\lim\limits_{n\to\infty}a_n=0\;.$

Another way to prove that $\;\lim\limits_{n\to\infty}a_n=0\;.$
From $(*)$ it follows that
$a_{n+1}<a_n\quad$ for all $\;n>\nu\;.$
Consequently, there exists $\;\lim\limits_{n\to\infty}a_n=\inf\limits_{n>\nu}\;\{a_n\}=l\ge0\;.$
If $\;l>0\;,\;$ then $\;\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}=\dfrac{l}{l}=1\;,\;$ but it is impossible because, for hypothesis, $\;\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}<1\;.$
Consequently, it results that $\;l=0\;$ that is
$\lim\limits_{n\to\infty}a_n=0\;.$
