# Ngô Quốc Anh

## September 30, 2007

### 1 bài tập khó về dãy số

Filed under: Các Bài Tập Nhỏ, Giải Tích 1, Giải Tích 4 — Ngô Quốc Anh @ 8:26

Suppose that $a_{n}$ and $b_{n}$ are two sequences of nonnegative numbers such that for some real number $N_{0}\geq 1$, the following recursion inquality holds: $a_{n+1}\le{a_{n}+b_{n}}$, for any $n\ge N_{0}$. Prove that if $\sum b_{n}<\infty$ , then $a_{n}$ is converges.

Solution. Since $\displaystyle\begin{gathered}a_{n+1}\leqslant a_{n}+b_{n}\hfill\\ a_{n+2}\leqslant a_{n+1}+b_{n+1}\leqslant a_{n}+\left({b_{n}+b_{n+1}}\right)\hfill\\ ...\hfill\\ \end{gathered}$

then $\displaystyle a_m \leqslant a_n + \sum\limits_{k = n}^{m - 1} {b_k } \leqslant a_n + \sum\limits_{k = n}^{ + \infty } {b_k }$

provided $m \geq n+1$. Now taking the $\limsup$ with respect to $m$ we have $\displaystyle \mathop {\lim \sup }\limits_{m \to + \infty } a_m - \sum\limits_{k = n}^{ + \infty } {b_k } \leqslant a_n$.

Now consider the $\liminf$ with respect to $n$ we deduce $\displaystyle\mathop {\lim \sup }\limits_{m \to + \infty } a_m \leqslant \mathop {\lim \sup }\limits_{m \to + \infty } a_m - \mathop {\lim \inf }\limits_{n \to + \infty } \sum\limits_{k = n}^{ + \infty } {b_k } \leqslant \mathop {\lim \inf }\limits_{n \to + \infty } a_n$

since $\mathop {\lim }\limits_{n \to + \infty } \sum\limits_{k = n}^{+ \infty } {b_k } = 0$. This and the fact that $\displaystyle \mathop {\lim \sup }\limits_{m \to + \infty } a_m \geqslant \mathop {\lim \inf }\limits_{n \to + \infty } a_n$

yields the conclusion.

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