Is the sum of an alternating series satisfies the following inequality

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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?

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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} $