TITLE: Disagreement with professor about Cauchy Sequence
QUESTION [7 upvotes]: Remark: in this post, all limits are infinite limits so I'll write just $\lim$ instead of $\lim_{n \rightarrow \infty}$ to save time and notation. Also I want to say that I already wrote the proof to the problem below using the definition but I want to discuss this particular proof.

Problem: If $\lim x_{2n} = a$ and $\lim x_{2n -1} = a$, prove that $\lim  x_n = a$.

Attempt:
Define $X_n = \{x_{2n}\} \cup \{x_{2n - 1}\}$. Take two elements from $X_n$ and let's calculate the limit of their difference:
$$ \lim (x_{2n} - x_{2n - 1}) = a - a = 0 $$
Therefore $X_n$ is a Cauchy sequence. Now we use the result that if a Cauchy sequence has a subsequence converging to $a$, then $\lim X_n = a$. End of proof.
Discussion:
I showed this proof to my professor. He said it's wrong because in the limit
$$ \lim_{n \rightarrow \infty, m \rightarrow \infty} x_n - x_m = 0$$
$n$ and $m$ cannot be related. I accepted his argument. Later, out of curiosity I looked up again the definition of Cauchy sequence and nowhere does it say that $n$ and $m$ cannot be related. Then I brought it up to my professor, and this time he said that I'm choosing a particular $n$ and $m$ and that I cannot do that in proving that $X_n$ is a Cauchy sequence.
I thought that I'm not fixing anything since $2n$ and $2n - 1$ are not fixed, and even then I could write
$$ \lim (x_{2n} - x_{2m}) = a - a = 0  $$
and what I wrote would still hold true..
However, I chose not to continue the discussion as I felt that if I insisted on this proof, my professor would feel antagonized and some negativity would be created. However I still fail to see why the proof is incorrect. Personally it felt like my professor just didn't accept that I came up with a clever proof.

REPLY [16 votes]: A sequence $(a_n)$ is a Cauchy sequence if and only if for every $\epsilon\gt 0$ there exists an $N\gt 0$ such that for all $n,m\geq N$, $|a_n-a_m|\lt \epsilon$.
Because the condition must be met by all pairs $n,m$, provided only that they be large enough, you cannot establish this condition by checking only those pairs in which, say, $m$ is a function of $n$, $m=f(n)$. 
What you wrote is insufficient: even if the difference between the $2n$th and the $(2n-1)$st term goes to zero, it is possible for the sequence to diverge (and hence not be Cauchy). 
For a simple example, consider the sequence with $a_{2n}=n$ and $a_{2n-1}=n$. That is, the sequence is $1,1,2,2,3,3,4,4,5,5,\ldots$. This sequence is not Cauchy, but it still satisfies that
$$\lim_{n\to\infty}(a_{2n}-a_{2n-1}) = 0.$$
More generally, you can always arrange for $|a_n-a_{f(n)}|$ to go to $0$, and yet for the sequence to diverge: just arrange for all those pairs to be equal,  and yet for $a_n$ to get arbitrarily large for sufficiently large $n$.

As to the original problem, here is one way to do it: let $\epsilon\gt 0$. We know that there is an $N$ such that if $2n\gt N$, then $|a_{2n}-a|\lt {\epsilon}$. And we know that there is an $M\gt 0$ such that if $2n-1\gt M$, then $|a_{2n-1}-a|\lt \epsilon$.
Take $K=\max{N,M}$. Then if $r\gt K$, then either $r$ is even, in which case $r=2n\gt K\geq N$, so $|a_r-a|\lt \epsilon$; or $r$ is odd, in which case $r=2n-1\gt K\geq M$, so $|a_r-a|\lt\epsilon$. Either way, $|a_r-a|\lt\epsilon$ for $r\gt K$. Thus, we conclude that the sequence converges to $a$.

REPLY [9 votes]: I think this example may get to the heart of the issue. Consider the truncated harmonic series $$H_n = \sum^n_{k=1} \frac 1 k.$$ We know that $\lim H_n =\infty$. However, we also see that $$H_{2n} - H_{2n-1} = \frac 1 {2n} \to 0.$$ To elaborate: clearly $H_n$ is not a Cauchy sequence, since it doesn't converge, and yet, if we fix $j \in \mathbb N$, we see that $$\lim_{n\to \infty} (H_{n+j} - H_{n}) = 0.$$ I believe this is what your professor means when he says that $n$ and $m$ cannot be related. To prove that $H_n$ is Cauchy, we need to fix $\epsilon > 0$ and find $N \in \mathbb N$ such that for all $n,m \ge N$, we have $$\lvert H_n - H_m \rvert < \epsilon.$$ As you have it written, it seems like you are fixing $\epsilon > 0$, finding $N \in \mathbb N$ and choosing a particular pair $n,m \ge N$ such that $\lvert H_n - H_m\rvert < \epsilon$, so the inequality will hold for this particular pair $n,m$ but will not hold for all $n,m \ge N$. 
If this doesn't help, I would still encourage you to apply your exact line of reasoning to $H_n$ and see if your reasoning implies convergence, because if so, there is some mistake in the reasoning. 
In general, I would say that any proof of this property that doesn't very critically invoke the fact that $x_{2n} \to a$ and $x_{2n-1} \to a$, is probably not the best proof. What I mean is: your proof should probably set $\epsilon > 0$ and explicitly take $N_1, N_2 \in \mathbb N$ such that for $n \ge N_1$ you have $\lvert x_{2n} - a\rvert < \epsilon$ and for $n \ge N_2$ you have $\lvert x_{2n-1} - a \rvert < \epsilon$, and work from there.