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It's true, that every convergent sequence is bounded, but is every bounded sequence convergent?

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Have you tried looking for an example of a bounded sequence that doesn't converge? –  Chris Eagle Jan 19 '13 at 15:05
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No. but every bounded real(or complex) sequence has convergent subsequence. –  tetori Jan 19 '13 at 15:09
    
And every bounded, monotonic sequence is convergent. –  Stefan Hansen Jan 19 '13 at 15:13
    
I would like to question the premise of the question as well. The sequence {1/n} with n in either the reals or the rationals is not bounded at n=0 and yet it converges as n -> Inf or n -> -Inf. –  DWin Jan 19 '13 at 18:59
    
@DWin, that is not a sequence. That is a net. See en.wikipedia.org/wiki/Net_(mathematics) –  MTS Jul 7 '13 at 18:55
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5 Answers

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No. For example, take the sequence $$a_n=\begin{cases} 0 & \text{ if }n\text{ is even}\\ 1 & \text{ if }n\text{ is odd} \end{cases}$$ It is bounded because it stays inside the interval $[0,1]$, but it has no limit.

Intuitively, you shouldn't expect that bounded $\implies$ convergent, because even if the terms of a sequence stay in some general area, doesn't mean that all of its terms must always be getting closer and closer to each other (which is what the notion of Cauchy sequence captures; a sequence in $\mathbb{R}$ or $\mathbb{C}$ is convergent $\iff$ it is Cauchy).

However, as user amWhy points out in their answer, every bounded sequence contains a convergent subsequence; in other words, we can pick out some terms of the sequence that are getting closer and closer to each other (even if they aren't getting closer to all the terms in the original sequence).

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If the sequence is $\mathbb{N}\longrightarrow\mathbb{R}$ instead of $\mathbb{N}\longrightarrow\mathbb{N}$, another good example would be: $a_n=\sin(n)$. –  MyUserIsThis Jan 19 '13 at 15:08
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@MyUserIsThis: This sequence is a sequence of real numbers. –  Asaf Karagila Jan 19 '13 at 15:19
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Hint: $\quad$Consider the sequence $\{a_n\},\;a_n = (-1)^n\,$

It is bounded in $[-1, 1]\; ($ indeed, $a_n \in \{-1, 1\}\; \forall a_n\in \{a_n\}),\;$ but $\;\lim_{n\to \infty} (-1)^n\;$ does not exist.

Note: it is true that every bounded sequence contains a convergent subsequence, and furthermore, every monotonic sequence converges if and only if it is bounded.


Added See the entry on the Monotone Convergence Theorem for more information on the guaranteed convergence of bounded monotone sequences.

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No. Take the sequence $a_n=i^n$ in the complex plane. It is bounded since it is contained in the circle $|z|\leq \sqrt{2}$. However it doesnt converge since its terms alternate depending on $n$ modulo 4.

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No, there are many bounded sequences which are not convergent, for example take an enumeration of $\mathbb Q\cap(0,1)$.

But every bounded sequence contains a convergent subsequence.

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Let $\{X_n\}$ be a convergent sequence, converging to $L$, then corresponding to $\epsilon=1$, there exist $N_0\in \Bbb N$ such that $|X_n-L|<1 \forall N\geqslant N_0$.

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