If $a_{n+1}\leq (1-\alpha_n)a_n+\alpha_n\beta_n,\;\;n=1,2,3,\cdots$, then prove that $a_n\to 0$ as $n\to\infty.$

Question

Good day all! Please, how do I prove this? I'm lost of ideas but I am sure that the solution is here. So, can anyone help me out?

Let $\{a_n\}$ be a sequence of non-negative real numbers such that $$a_{n+1}\leq (1-\alpha_n)a_n+\alpha_n\beta_n,\;\;n=1,2,3,\cdots$$ where $\{\alpha_n\}\subset [0,1],\{\beta_n\}\subset [0,1]$ and $\sum\alpha_n=\infty$, $\lim\beta_n=0$. Prove that $a_n\to 0$ as $n\to\infty.$

Answer 1

If $a_n\leq b_n$ then $a_{n+1}\leq (1-\alpha_n)b_n+\alpha_n b_n=b_n$, hence inductively $a_m\leq b_n$ for all $m\geq n$.

So, if there are infinite many $b_i$ such that $a_{n_i}\leq b_i$ for some $n_i$ then $a_n$ converges.

Therefore, assume that there is $M$ such that $a_n>b_n$ for all $n\geq M$. Then, we have $a_{n+1} \leq (1-\alpha_n)a_n +\alpha_na_n=a_n$. Hence, $a_n$ converges to $a$.

Assume that $a\neq 0$.

Since $b_n$ converges to zero we can find $N$ such that $\beta_n-a_n \leq -\frac{a}{2}$ for all $n \geq N$.

For $n\geq N$, telescoping $a_{n+1}\leq (1-\alpha_n)a_n+\alpha_n\beta_n$ we obtain \begin{align*} a_{n+1}-a_N&\leq\sum_{k=N}^n \alpha_k(\beta_k-a_k)\\ &\leq -\frac{a}{2}\sum_{k=N}^n \alpha_k \longrightarrow -\infty \end{align*} A contradiction since the LHS is $a-a_N$.