limit of the sequence $s_n=\sum_{k=1}^n (-1)^{k+1}a_k$ Consider the sequence $s_n=\sum_{k=1}^n (-1)^{k+1}a_k$, with $a_k$ for every $k\geq1$, with $\lim_{k\rightarrow\infty}a_k=0$. Suppose that the sequence $s_{2n+1}$, $n\geq0$, is decreasing and the sequence $s_{2n}$, $n\geq1$, is increasing. Prove that $s_n$ converges.
Before this exercise, I have proved the following result

If $x_n$ is increasing, $y_n$ is decreasing with $\lim (y_n-x_n)=0$ and, for every $n\in\mathbb{N}$, $y_n-x_n\geq0$, then both the sequences converge for the same limit.

I was thinking that I can use this one, because $x_n=s_{2n}$ is increasing and $y_n=s_{2n+1}$ is decreasing and $x_n-y_n\geq0$ and 
$$\lim(y_n-x_n)=\lim(s_1+s_3+\cdots)+\lim(s_2+s_4+\cdots)=\lim s_n$$
then $s_n$ converges.... Do you agree with me?
 A: You have the series
$$s_n=\sum_{k=1}^n (-1)^{k+1}a_k \tag{1}\label{eq1A}$$
Note by "decreasing" I assume it means $\le$ and by "increasing" I assume it means $\ge$. If it's to be considered "strict" instead, then replace the $\le$ with $\lt$, and $\ge$ with $\gt$ below.
You have
$$\lim_{k\rightarrow\infty}a_k=0 \tag{2}\label{eq2A}$$
Since the sequence $s_{2n + 1}$, $n \geq 0$, is decreasing, you thus have that
$$\begin{equation}\begin{aligned}
s_{2n+3} - s_{2n+1} & \le 0 \\
\sum_{k=1}^{2n+3} (-1)^{k+1}a_k - \sum_{k=1}^{2n+1} (-1)^{k+1}a_k & \le 0 \\
(-1)^{(2n+3) + 1}a_{2n+3} + (-1)^{(2n+2) + 1}a_{2n+2} & \le 0 \\
a_{2n+3} - a_{2n+2} & \le 0 \\
a_{2n+2} & \ge a_{2n+3}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Similarly, since the sequence $s_{2n}$, $n\geq1$, is increasing. you get
$$\begin{equation}\begin{aligned}
s_{2n+2} - s_{2n} & \ge 0 \\
\sum_{k=1}^{2n+2} (-1)^{k+1}a_k - \sum_{k=1}^{2n} (-1)^{k+1}a_k & \ge 0 \\
(-1)^{(2n+2) + 1}a_{2n+2} + (-1)^{(2n+1) + 1}a_{2n+1} & \ge 0 \\
-a_{2n+2} + a_{2n+1} & \ge 0 \\
a_{2n+1} & \ge a_{2n+2}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Using $n = 0$ in \eqref{eq3A} gives
$$a_2 \ge a_3 \tag{5}\label{eq5A}$$
while using $n = 1$ in \eqref{eq4A} gives
$$a_3 \ge a_4 \implies a_2 \ge a_3 \ge a_4 \tag{6}\label{eq6A}$$
Now, repeating the procedure using $n = 1$ in \eqref{eq3A} and $n = 2$ in \eqref{eq4A}, and combining the results with \eqref{eq6A}, gives
$$a_2 \ge a_3 \ge a_4 \ge a_5 \ge a_6 \tag{7}\label{eq7A}$$
You can fairly easily prove, such as by using induction (which I'm leaving to you to do), that
$$a_k \ge a_{k + 1}, \; \forall \; k \ge 2 \tag{8}\label{eq8A}$$
This means the $a_k$ terms, apart from possibly $a_1$, are all monotonically decreasing. Thus, using \eqref{eq2A}, the alternating series test states that $s_n$ converges.

Note in your proof attempt, if you let $x_n = s_{2n}$ and $y_n = s_{2n+1}$, then $x_n - y_n = -(-1)^{(2n + 1) + 1}a_{2n + 1} = -a_{2n + 1}$, which is actually $\le 0$ instead as you have $x_n$ and $y_n$ mixed around. Also, in your next line where you state
$$\lim(y_n - x_n) = \lim(s_1 + s_3 + \cdots) + \lim(s_2 + s_4 + \cdots) = \lim s_n$$
you seem to use that $y_n$ and $x_n$ are the sums of $s_{2n + 1}$ and $s_{2n}$ from $1$ up to $n$ rather than just being those terms as you earlier defined them.
Nonetheless, with your definitions, you have $y_n - x_n = a_{2n + 1}$, so $\lim_{n \to \infty}(y_n - x_n) = 0$. Also, as shown in \eqref{eq8A}, all $a_{k}$ for $k \ge 2$ are non-negative (as it's a monotonically decreasing series with a limit of $0$) so for every $n \in \mathbb{N}$, $y_n - x_n \ge 0$. Thus, you could also use the result which you stated you proved earlier.
