Proving that the alternating sequence is Cauchy Prove using the definition of a Cauchy sequence that X is Cauchy
$$
X = \cfrac{(-1)^n}{(2n+1)}
$$
thank you.
If we break this into two cases where in case 1. n is odd, case 2. n is even, will it be valid to say "since these two sequences are Cauchy, then the above is Cauchy"?
 A: The argument of using the sequences of even and odd terms does not work. For example, consider the sequence of real numbers given by $x_n=0$ if $n$ is even, and $x_n=1$ if $n$ is odd. Then the sequences of even and odd terms are both Cauchy, since they are constant, but the sequence as a whole is not, since it is not convergent.
A solution for your problem would be, for example, showing that $X$ converges, and for this you can argue that the sequences of even and odd terms both converge to the same limit.
EDIT: After OP's request, here is the proof that the sequence is Cauchy directly from the definition, without using the fact that convergent sequences are Cauchy.
Given $\varepsilon>0$, we need to show that there exists a $n_0 \in \mathbb{N}$ such that $\lvert a_n-a_m \rvert < \varepsilon$ for any $n,m \geq n_0$. In this case
$$\lvert a_n-a_m\rvert = \left\lvert\frac{(-1)^n}{2n+1} + \frac{(-1)^m}{2m+1}\right\rvert \leq \frac{1}{2n+1}+\frac{1}{2m+1} < \frac{1}{n}+\frac{1}{m},$$
so, taking $n_0 > \varepsilon/2$ we get that if $n,m \geq n_0$, then $\lvert a_n-a_m \rvert < \varepsilon$, which concludes the proof.
A: Notice that the sequence is convergent to zero. As you are supposed to prove by definition, you prove that for any $\epsilon >0$, there exists a natural number $N$ such that $|x_n| < \frac{\epsilon}{2}  $ whenever $n \geq N$, just do not mention that $lim_{n \rightarrow \infty} x_n =0$. Obviously the argument holds for $\{x_m\}$ as well. Now, $|x_n - x_m| \leq |x_n| + |x_m| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2}$ and so on.
A: If $a_n$ is Cauchy and $b_n$ is Cauchy then is the sequence $c_n = \begin{cases} a_n, \ n \text{ even} \\ b_n, \ n \text{ odd} \end{cases}$ Cauchy?  See other peoples' answers.  
However, $|c_n - c_m| \leq \max\{|a_n - a_m|, |b_n - b_m|, |a_n - b_m|, |b_n - a_m|\}$.
Clearly the first two are $\lt \varepsilon$ eventually.  The third one is:
$$
a_n - b_n = \dfrac{1}{2n+1} - \dfrac{-1}{2m + 1} = \dfrac{2m + 1 - 2n - 1}{(2m + 1)(2n+1)}
$$
which looks essentially like $\dfrac{m - n}{mn}$.  Prove the Cauchy quality of that. Namely, that for any $\varepsilon \gt 0$ there is a natural $N$ such that for all $m,n \gt N$ you have that it's absolute value is $\lt \varepsilon$.  
Thus, choose $\varepsilon \gt 0$, and take the maximum of the $N_i, \ i = 1..4$ needed to Cauchy out the $4$ absolute values.  Then $|c_n - c_m| \lt \varepsilon$.  
So your guess is partially correct.  Except you need an additional lemma.
Def.  Define two sequences $a_n, b_n$ to be cross-Cauchy if together they are Cauchy as in the above (namely, $|a_n - b_m|$ in place of $|a_n - a_m|$ in the definition of Cauchy).
Lemma.  If $a_n, b_n$ are both Cauchy and cross-Cauchy with each other, then the sequence constructed by interleaving the two sequences is a Cauchy sequence.
$\square$
