# How to prove that a sequence is convergent?

Let $\{u^k\}\subset \mathbb{R}^n$ be a sequence satisfying that there exists a subsequence $\{u^{k_i}\}\subset \{u^k\}$ converging to $\bar{u}\in \mathbb{R}^n$. I would like to ask when we have a stronger conclusion that $\{u^k\}$ converges to $\bar{u}$. For example, if $\{\|u^k-\bar{u}\|\}$ is monotonically decreasing then $\{u^k\}$ converges to $\bar{u}$.

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A Cauchy sequence converges if and only if it has a convergent subsequence. –  Arturo Magidin Jul 17 '12 at 5:56
Dear Arturo Magidin. Since $\{u^k\}\subset\mathbb{R}^n$, $\{u^k\}$ is a Cauchy sequence iff it is convergent. –  blindman Jul 17 '12 at 6:02
Are you asking that, if a sequence has a convergent subsequence, then the main sequence converges? –  Mhenni Benghorbal Jul 17 '12 at 6:09
@Mhenni Benghobal. I would like to find some sufficient conditions for the sequence $\{u^k\}$ to be convergent in the case $\{u^k\}$ has a convergent subsequence. –  blindman Jul 17 '12 at 6:16

Let $(X,d)$ be a metric space, and let $(a_k)$ be a Cauchy sequence in $X$. Then $(a_k)$ converges iff $(a_k)$ has a convergent subsequence.
in our case $\{u^k\}\subset \mathbb{R}^n$, we don't consider in a general metric space. –  blindman Jul 17 '12 at 6:33
Yes, but since $\mathbb{R}^n$ is in fact a metric space, the result applies. –  KReiser Jul 17 '12 at 6:39
@KReiser, I understand completely what you mentioned, but I would like to find more properties of the sequence $\{u^k\}\subset \mathbb{R}^n$ to guarantee $\{u^k\}$ to be convegent in the case it has a convergent subsequence. –  blindman Jul 17 '12 at 6:54