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Given a sequence $(a_n)$ prove that

a) If it is known that $|a_{n+1}-a_n|< 1/n$ for all $n$, show that $(a_n)$ need not be a Cauchy Sequence.

b) If it is known that $|a_{n+1}-a_n| < 1/2^n$ for all $n$, show that $(a_n)$ is a Cauchy Sequence.

Do I prove it using the definition of Cauchy sequence i.e for a) I've tried to find $\epsilon$ such that $|a_{n+1}-a_n| > \epsilon$ but failed.

Anyone can help?

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1  
You do not need $|a_{n+1}-a_n|>\varepsilon$, but $|a_p-a_q|>\varepsilon$ for some $p,q$ going to infinity. –  zozoens Feb 3 at 10:37
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The problem is not formulated correctly. Let me try to fix it. –  Andrej Bauer Feb 3 at 10:43
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See this for a). –  David Mitra Feb 3 at 11:01
    
Check this steps to reach the answerAnswer. –  Semsem Feb 3 at 11:11

2 Answers 2

I suspect there is confusion about understanding the problem. The problem is really asking you whether you are allowed to conclude that $(a_n)$ is Cauchy if all you know is the inequality as stated.

Thus, in order to solve (a) you must find a particular concrete sequence which satisfies the condition but is not Cauchy. In order to solve (b) you must show that every sequence satisfying the condition is Cauchy. So, here are the solutions:

(a) We must give a sequence $(a_n)$ such that $|a_{n+1} - a_n| < 1/n$ and the sequence $(a_n)$ is not Cauchy. Here is one: define $a_n = 1 + 1/2 + \cdots + 1/n$. Then for all $n$ we have $$|a_{n+1} - a_n| = 1/(n+1) < 1/n,$$ however, as is well known the sequence $(a_n)$ is divergent (it is known as the harmonic series).

NB: There are sequences which satisfy $|a_{n+1} - a_n| < 1/n$ but are Cauchy. For instance, the sequence $a_n = 0$ is Cauchy and $|a_{n+1} - a_n| = |0 - 0| = 0 < 1/n$.

(b) We must show that every sequence $(a_n)$ which satisfies $|a_{n+1} - a_n| < 1/2^n$ is Cauchy. Proof: suppose $(a_n)$ is a sequence such that $|a_{n+1} - a_n| < 1/2^n$ for all $n$. To prove that $(a_n)$ is Cauchy, consider any $\epsilon > 0$. We must find $N$ (depending on $\epsilon$ only) such that $|a_m - a_n| < \epsilon$ for all $n, m$ satisfying $m \geq n \geq N$. We claim that $N = 1 + \log_2 \epsilon$ is good. As in @HeeKwonLee's answer, we may compute that for all $m \geq n \geq N$ we have $$|a_m - a_n| < 1/2^{n-1} \leq 1/2^{N - 1} = 1/2^{\log_2 \epsilon} = \epsilon$$

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a) $a_n = \sum_{k=1}^n\frac{1}{2k}$ so that it satisfies the condition. To prove that $\{ a_n\}$ is not Cauchy, what must we do ?

b) $$|a_{m}-a_n| \leq \sum_{k=n+1}^m |a_k-a_{k-1}| = \sum_{k=n+1}^m \frac{1}{2^{k-1}} <\frac{1}{2^{n-1}} $$ Can you find an $\epsilon$ ?

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