prove the limit of a decreasing sequence prove that if a sequence $\{a_n\}$ is decreasing and there exists a subsequence $\{a_{n_k}\}$ so that $\lim_{k\to\infty} \{a_{n_k}\} =L$ then $\lim_{n\to\infty} \{a_n\}=L$.
I don´t know how to do this can someone help me please.
 A: Hint.  You need to do two things: 
(1) prove that $\{a_n\}$ has a limit;
(2) prove that this limit is $L$.
You have probably had a theorem in your course: "if a sequence converges to a limit, then any subsequence converges to the same limit".  Together with the information you are given, this makes part (2) easy.
For part (1), a decreasing real sequence which is bounded below must converge to a limit.  To prove in your situation that every $a_n$ is greater than or equal to $L$, assume not.  Then some term $a_n$ and every subsequent term (why?) is less than $L$, which is impossible because. . .
See if you can complete the argument from here.
A: Well a decreasing sequence either tends to a limit or tends to $-\infty$. Now if $a_{n} \to -\infty$ then every subsequence also does so which is not the case here. Hence $a_{n}$ does tend to a limit. Then every subsequence must tend to the very same limit. Since $a_{n_{k}}$ tends to $L$ it follows that the original sequence $a_{n}$ also tends to $L$.
The basic fact to be understood is that if any sequence (whether monotonic or not) tends to a limit $L$ (or to $\pm\infty$) then every subsequence also does so. Only when a sequence is oscillating we can have a scenario that difference sub-sequences have different behavior. In the current question the monotonic nature of the sequence $a_{n}$ forbids any oscillation.
