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Can someone provide me the definition of a (finite ) $\lim (s_n)$ and how it correlates to the definition of $\limsup(s_n)$?

$\lim(s_n)=+\infty$ if $\forall M>0, \exists N=N(M)\in \Re$ s.t.$\forall n>N $ we have $s_n >M$.

I believe this is the definition for the infinite case....

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If the limit exist, then $\limsup s_n=\liminf s_n=\lim s_n$. – Carlos Eugenio Thompson Pinzón Nov 6 '13 at 1:09
up vote 1 down vote accepted

Let $s_n$ be a real sequence. Let $b_n$=sup{$s_n,s_{n+1},....$} for every $n=1,2,...$. Then $\limsup(s_n)=\overline {\lim s_n}=\inf b_n=(\mathop{\mathrm{inf\,sup}}s_k)$ for $k\geq n$. To understand it better if we have a sequence $s_n=(-1)^n$ then $\limsup s_n=1$ and $\liminf s_n=-1$. Also if $\limsup s_n=s$ then this means that for every $ε>0$ the set {$n:s+ε<s_n$} is finite.

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Take $s_n=0$ if $n$ is an even prime, and $s_n=(1+\frac1n)$ otherwise then $s=\limsup s_n=1$ yet there are infinite elements of $s_n$ larger than $s$, and finite elements of $s_n$ smaller than $s$. – Carlos Eugenio Thompson Pinzón Nov 6 '13 at 1:28
you are right.i corrected it – Haha Nov 6 '13 at 1:32

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