May 19th, 2017, 07:54 PM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Sequences
Consider the following two statements : 1. If $(a_n)$ is any real sequence, then $\displaystyle \frac{a_n}{1+a_n} $ has a convergent sub sequence. 2. If every sub sequence of $(a_n)$ has a convergent subsequence, then $(a_n)$ is bounded. A) Both 1 & 2 are true. B) Both 1 & 2 are false. C) 1 is false but 2 is true. D) 1 is true but 2 is false. I have checked (1) and it is not true in my case where $\displaystyle \sum_{n=0}^{\infty} 2^n$ is the sub sequence. (2) is true if every sub sequence of all sub sequences of $(a_n)$ is convergent then all sub sequences of $(a_n)$ will be convergent. That implies $(a_n)$ is convergent then it should be monotonic and bounded. Correct me If I'm wrong 
May 20th, 2017, 07:09 PM  #2  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
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May 20th, 2017, 08:11 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,551 Thanks: 2554 Math Focus: Mainly analysis and algebra 
I think that your answer to 1) is incorrect. For large positive values of $a_n$, $\frac{a_n}{1+a_n} \approx 1$. For small values of $a_n$, $\frac{a_n}{1+a_n} \approx 0$. For large negative values of $a_n$, $\frac{a_n}{1+a_n} \approx 1$. In fact, $\frac{a_n}{1+a_n}$ is bounded above and below, so it can't diverge unless it oscillates. In either case it must have convergent subsequences. Note that this doesn't mean that every subsequence is convergent, but at least one of them is. 

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