alternate proof to alternating series test Prove the alternating series test from the Cauchy convergence theorem for series. We want to show $\sum_{k=1}^{\infty} (-1)^{k+1}a_k$ converges. 
Proof:
Let $\epsilon>0$. Let $k,k'\in\mathbb{N}$ and let $a_k$ be a decreasing monotone sequence of non negative  numbers such that $a_k\rightarrow 0$. It follows by definition of convergence that there exists a $N'\in\mathbb{N}$ such that if $k\geq N'$ then $|a_k|=a_k<\frac{\epsilon}{k'}$. Now let $N=N'$ such that if $n>N'$ then $|(-1)^{n+2}a_{n+1}+....+(-1)^{n+k'+1}a_{n+k'}|\leq |(-1)^{n+2}a_{n+1}|+|(-1)^{n+3}a_{n+2}|+...+|(-1)^{n+k'+1}a_{n+k'}|=a_{n+1}+..+a_{n+k'}\leq (k')(a_{n+1})<k'*\frac{\epsilon}{k'}=\epsilon$. Thus the series converges. Is this correct? The only issue I could see is the $k'$ but since its fixed I'm not sure. I just want to know if my proof is correct or what can i do to correct it?
 A: Some intuition why your proof isn't quite correct: When you used the triangle inequality
$$|(-1)^{n+2}a_{n+1} + \cdots + (-1)^{n+k'+1}a_{n+k'}| \leq |(-1)^{n+2}a_{n+1}| + \cdots + |(-1)^{n+k'+1}a_{n+k'}|\\
 = a_{n+1} + \cdots + a_{n+k'} $$
you removed the alternating nature of the series, so if your proof was correct, you would have proved convergence for the non-alternating series $\sum a_k$ as well. 
The technical trouble comes from the fact that to show Cauchy, you need to be able to select any $n, m \geq N$ ($m$ and $n$ can not depend on each other, for example), but by selecting $k'$ beforehand and then letting $N$ depend on $k'$, you force $m = n+k'$ to depend on $n$ through your selection of $N$. (Does this make sense?). 
Here's an argument to help you prove what you want. Let $b_k = (-1)^k$ and let's note that for any $m \leq n$ $$\Big| \sum_{k=m}^n b_k \Big|  \leq 1$$ 
From a telescoping series argument, you can show that $a_k = \sum_{l=m}^{k-1} (a_{l+1} - a_{l}) + a_m$. Therefore,
$$
\sum_{k=m}^n b_k a_k   = \sum_{k=m}^n b_k \Big( \sum_{l=m}^{k-1} (a_{l+1} - a_l) + a_m \Big) = \sum_{l=m}^{n-1} \Big\{ (a_{l+1} - a_l) \sum_{k=l}^n b_k \Big\} + a_m \sum_{k=m}^n b_k
$$
You can now deduce that 
$$
\Big| \sum_{k=m}^n b_k a_k \Big| \leq \sum_{l=m}^{n-1} \Big\{ |a_{l+1} - a_l| \Big|\sum_{k=l}^n b_k \Big| \Big\} + a_m \Big| \sum_{k=m}^n b_k \Big| \leq \sum_{l=m}^{n-1}|a_{l+1} - a_l| + a_m.
$$
Since $a_l \geq a_{l+1}$ for every $k$, $|a_{l+1} - a_l| = a_l - a_{l+1}$, which reduces our last inequality to
$$
\Big| \sum_{k=m}^n b_k a_k \Big| \leq \sum_{l=m}^{n-1} (a_{l} - a_{l+1}) + a_m = 2a_m - a_n.
$$
where the last equality used another telescoping series argument to show $\sum_{l=m}^{n-1} (a_{l} - a_{l+1}) = a_m - a_{n}$. 
Now, had you originally chosen $N \in \Bbb{N}$ with $n, m \geq N$, then (by monotonicity) $a_m \leq a_N$ and also $a_n \geq 0$, so you finally arrive at
$$
\Big| \sum_{k=m}^n b_k a_k \Big| \leq 2a_m - a_n  \leq 2 a_N -  0 = 2a_N.
$$
So, at last, if $\epsilon  > 0$ and you choose $N$ such that $a_N < \epsilon /2$, what all this work has shown is that if $N \leq m, n$ (and w.l.o.g. $m < n$) then
$$
\Big| \sum_{k=m}^n b_k a_k \Big| \leq 2a_N < 2(\epsilon/2) = \epsilon.
$$
A: Here is one way to prove an alternating series is Cauchy (note this is just an adaptation of the usual proof, as mentioned in robjohn's comment, above that an alternating series converges).
Let $(a_n)$ be a decreasing sequence of nonnegative numbers.
We have, for $k$ even,
$$
a_n-a_{n+1}+ a_{n+2}+\cdots+ a_{n+k}
=a_n+(a_{n+2}-a_{n+1})+\cdots+(a_{n+k}-a_{n+k-1})\le a_n
$$
and 
$$
a_n-a_{n+1}+ a_{n+2}+\cdots+ a_{n+k}
= (a_n-a_{n+1})+ \cdots+ (a_{n+k-2}-a_{n+k-1})+a_{n+k}\ge0.
$$
A similar argument shows 
$$0\le a_n-a_{n+1}+\cdots +(-1)^k a_{n+k}\le a_n
$$
for $k$ odd.
It then follows that 
$$\tag{1}
\biggl|\sum_{i=n}^{n+k}(-1)^i a_i\biggr|\le a_n
$$
for all admissable $n,k$ (note $|\sum_{i=n}^{n+k}(-1)^i a_i = |- \sum_{i=n}^{n+k}(-1)^i a_i| $).
It follows from $(1)$ that $\sum_{i=1}^\infty (-1)^i a_i$ is Cauchy if, in addition to the above assumptions, $(a_n)$ converges to $0$. 
