If a sequence $\{a_n\}$ satisfies the Inequality $a_{n+1} < ka_{n}$, then show that $ \lim\limits_{n \to \infty} a_n =0$ where $0< k , a_n< 1$ I know one solution.

Consider $\sum a_n$

Then use ratio test to show that the series converges, hence the sequence.

Any other Ideass !
I will suppose $a_n \ge 0$, as otherwise the statement is wrong.

By induction, we have $a_n &lt; k^na_0$, hence $0 \le a_n \le k^n a_0$. As $0 \to 0$ and $k^n a_0 \to 0$ (we have $k \in (0,1)$), by the squeeze theorem $a_n \to 0$.
we have a theorem if a sequence is bounded and descend this sequence ins convergence and this condition is true for your question .and i think we have $a_n\geq 0 $