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}


\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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