KingOliver
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 Aug 16 accepted Find a sequence $\{a_n\}$ that converges to $a$ without getting “closer and closer” Aug 16 asked “So That” vs. “Such That” Aug 15 accepted Limit point versus limit and implications for convergence Aug 15 comment Limit point versus limit and implications for convergence So that sequence I have constructed works as a counterexample then. Thanks @DavidMitra Aug 15 comment Limit point versus limit and implications for convergence Does a sequence of this form "count"?: $(a_n) = 1$ for $n = 2k, k \in \mathbb N$, $(a_n) = n$ for $n = 2k - 1, k \in \mathbb N$? Aug 15 asked Limit point versus limit and implications for convergence Aug 15 accepted Proving that a sequence is Cauchy Aug 15 comment Proving that a sequence is Cauchy Thanks! I see the convergent series, which gives me the $N$ I need to pick. Aug 15 asked Proving that a sequence is Cauchy Aug 14 comment Counterexample Check for Sum of Limit Points of Subsequences Ah ok. Thank you all for your help, I am self-studying introductory real analysis and slowly getting the hang of the material. Aug 14 comment Counterexample Check for Sum of Limit Points of Subsequences Wondering @BrianM.Scott Aug 14 accepted Counterexample Check for Sum of Limit Points of Subsequences Aug 14 comment Counterexample Check for Sum of Limit Points of Subsequences @Rolando posted what looked like a proof of this conjecture, and then deleted it after you posted your answer. It looked something like this: Let $\epsilon >0$ be given. Since $a_n \to c$ and $b_n \to d$, there are $N_1, N_2$ such that for $l \ge N_1$, $k \ge N_2$ $|a_l - c| \le \epsilon/2$ and $|b_k - d| \le \epsilon/2$. Let $m = \max\{N_1, N_2\}$ and therefore we have for $n \ge m$ we have $|a_n - c + b_n - d| \le \epsilon$. Is the problem with his proof that he is not considering subsequences, and that he interpreted the limit points as limits? Aug 14 asked Counterexample Check for Sum of Limit Points of Subsequences Aug 14 accepted Convergent Subsequence in $\mathbb R$ Aug 14 comment Convergent Subsequence in $\mathbb R$ Never mind, of course you are right, because the denominator $\to \infty$ so the sequence $\to 0$, sorry I lost perspective. Thanks again! Aug 14 comment Convergent Subsequence in $\mathbb R$ Is it fair to say that since those subsequences are the only ones I have to consider because they are the only two convergent subsequences? Other than ones like $\langle a_{4k} \rangle$ which are contained within the one considered? Aug 14 asked Convergent Subsequence in $\mathbb R$ Aug 14 comment Proving that $\mu$ is $\sup S$ @JayeshBadwaik then why the $\Longleftrightarrow$? Aug 14 comment Proving that $\mu$ is $\sup S$ Thank you. This is clear to me now @PeterTamaroff