Let $\{a_k\}_{k\in \mathbb{N}}, \{b_k\}_{k \in \mathbb{N}}, $ and $\{c_k\}_{k \in \mathbb{N}}$ be sequences of non-negative real numbers satisfying the inequality,
\begin{align}
a_{k+1} \leq a_k – b_k + c_k
\end{align}
for $k=0,1,2,\cdots$ and $\sum_{k=0}^{\infty} c_k < \infty$.
I want to show that $b_k \rightarrow 0$ and $\{a_k\}_{k \in \mathbb{N}}$ has a limit as $k \rightarrow \infty$.
Naively due to $\sum_{k=0}^{\infty} c_k < \infty$, we know $\lim_{k \rightarrow \infty} c_k =0$.
So taking $k \rightarrow \infty$, one have
\begin{align}
a_{\infty+1} \leq a_{\infty} – b_{\infty} + c_{\infty}
\end{align}
and since $c_{\infty} =0$ and if limit of $a_k$ exists this gurantees that $\lim_{k\rightarrow \infty} b_k =0$. But we do not have any information of $a_k$ except that inequality.
How one can prove the above theorem?
Best Answer
Let $n <m$. Then $\sum\limits_{k=n}^{m} (a_{k+1}-a_k) \leq \sum\limits_{k=n}^{m} c_k \to 0$ as $m >n \to \infty$. The left side is a telescopic sum and you get $a_{m+1}-a_n \leq \sum\limits_{k=n}^{m} c_k$. Clearly, this implies that $(a_n)$ is bounded. Suppose $a$ and $b$ are two limit points of $(a_n)$. Then we get $a-b \leq 0$ and $b-a \leq 0$ by choosing $n$ and $m$ appropriately. [I will leave the details to you]. It follows that $(a_n)$ is convergent. Now $b_k \leq c_k -(a_{k+1}-a_k) \to 0$.