Is the sum of an alternating series satisfies the following inequality 
Question if $s$ is sum of the alternating series $\sum (-1)^{n+1}z_n$ and if $s_n$ is the nth partial sum then, $|s-s_{n}|≤z_{n+1}$

My attempt:
$|s-s_n|=|s-s_{n+1}+s_{n+1}-s_n|$
$$≤|s-s_{n+1}|+|s_{n+1}-s_n|$$
$$≤\epsilon + z_{n+1}$$
($s_n\rightarrow s$ so that, $s_{n+1}\rightarrow s$)
Now as $\epsilon >0$ is arbitrary,  letting $\epsilon\rightarrow 0$,  we havve $|s-s_{n}|≤z_{n+1}$
Is my attempt is correct? Please help....
Is there is any other way?
 A: Here, two cases may arrise,one is when " $(z_n)$ is not monotone " and another is "$(z_n)$ is monotone ".
Case $1$, when $(z_n)$ is not monotone, then your statement is wrong.
Take the example, $z_n = \begin{cases}2^{-n} & \text{ n is even} \\ 3^{-n} & \text{ n is odd }\end{cases}$. ,
The sum, $\sum_{n=1}^{\infty} (-1)^{n+1} z_{n} = \frac{1}{24} $.
Clearly, $|s - s_1|=| \frac{1}{24} - \frac{1}{3} | = \frac{21}{72} > \frac{1}{4} = |z_2| $.
Case $ 2 $ , when $(z_n) $ is monotone,
Since,  $ \sum_{k=1}^{\infty} (-1)^{k+1} z_{k} = s $,
Just do, $(s - s_{n}) = \sum_{k=1}^{\infty} (-1)^{k+1} z_{k} -\sum_{k=1}^{n} (-1)^{k+1} z_{k}   =\pm z_{n+1} \mp( ( z_{n+2} - z_{n+3}) + (z_{n+4} - z_{n+5})+( z_{n+6} - z_{n+7}) + ............................      ) $ .
If $(z_{n}) $ is monotone,then , the sum = $( ( z_{n+2} - z_{n+3}) + (z_{n+4} - z_{n+5})+( z_{n+6} - z_{n+7}) + .........) $
must be positive ,
Now take          $( ( z_{n+2} - z_{n+3}) + (z_{n+4} - z_{n+5})+( z_{n+6} - z_{n+7}) + .........) =A $
A is some positive real.
Now,  $(s - s_{n}) = \pm z_{n+1} \mp A = \pm(z_{n+1} - A) $.
Hence, $|s - s_{n}| \le z_{n+1} $
