A sequence $\{a_n\}$ is such that $\lim_n(a_{n+1} - a_n) = 0$. Given some additional properties of $a_n$ prove that it converges. 
Let $\{a_n\}, n\in\Bbb N$ be a sequence such that:
  $$
\lim_{n\to\infty}(a_{n+1} - a_n) = 0\tag1
$$
  The sequence also satisfies the following properties:
  $$
\begin{align*}
|a_{n+2}-a_{n+1}| \le |a_{n+1}-a_{n}|\tag 2\\
(a_{n+2} - a_{n+1})(a_{n+1}-a_n) \le 0 \tag 3
\end{align*}
$$
  Prove $\{a_n\}$ converges.

While looking at the problem first time I was thinking that $(1)$ implies convergnce of $\{a_n\}$, but looking more carefully one may think of a harmonic series:
$$
a_n = \sum_{k=1}^n{1\over k}
$$
Which is clearly divergent as $n\to\infty$, even though $(1)$ still holds. So obviously there is some extra work to do. So below are some of my further findings. First, let's define a new sequence:
$$
b_n = a_n - a_{n-1}\tag 4
$$
If a sequence converges then its absolute values also converge:
$$
\lim_{n\to\infty}b_n = 0 \implies \lim_{n\to\infty}|b_n| = 0
$$
If we now take a look at $(3)$ there are two possible cases for that statement to hold, either:
$$
b_{n+2} \ge 0\\
b_{n+1} \le 0 \tag5
$$
or:
$$
b_{n+2} \le 0\\
b_{n+1} \ge 0 \tag6
$$
But at the same time:
$$
b_{n+2}b_{n+1} \le 0\\
b_{n+1}b_{n} \le 0\\
b_{n}b_{n-1} \le 0\\
\cdots
$$
Let's stick to case $(5)$ for definiteness. This implies:
$$
0 \le b_{n+2}\le b_n \le b_{n-2} \le \cdots \tag7
$$
At the same time since $|b_n|$ is monotically decreasing and $b_{n+1}\le 0$:
$$
\cdots \le b_{n-3} \le b_{n-1} \le b_{n+1} \cdots \le 0 \tag8
$$
It looks like $b_n$ is an alternating sequence, which consists of two subsequences both convergent to $0$, that is because $(7)$ is monotonically decreasing towards $0$ and $(8)$ is monotonically increasing towards $0$. And this depends on whether we assume $(5)$ or $(6)$. 
That is where I got stuck. I do not see how to combine those finding to show that $a_n$ is actually convergent. How do I proceed?
 A: It is convergence of $\sum b_n$ (not that of $(b_n)$) that gives convergence of $(a_n)$. Apply Alternating Series Test to see that the series $\sum b_n$ is convergent. Then note that $b_1+b_2+...+b_n=a_1-a_n$. Hence $\lim a_n$ exists.  
A: Ok, following @Kavi Rama Murthy's suggestions here is what I eventually got (hopefully correct). 
I'm still going to stick to how $b_n$ is defined:
$$
\lim_{n\to\infty}b_n = 0\\
|b_{n+2}| \le |b_{n+1}|\\
b_{n+2}b_{n+1} \le 0
$$
Let's now define a partial sum $S_n$:
$$
S_n = \sum_{k=1}^n (-1)^{k-1}|b_k|
$$
Here the $(-1)^{k-1}$ actually depends on which terms are positive and which are negative (which is defined by eiher $(5)$ or $(6)$ in the OP). Consider two partial sums for odd and even numbers of terms:
$$
\begin{align*}
S_{2p+1} = \sum_{k=1}^{2p+1}(-1)^{k-1}|b_k| \\
S_{2p} = \sum_{k=1}^{2p}(-1)^{k-1}|b_k|
\end{align*}
$$
Consider the sum of odd amount of terms:
$$
\begin{align}
S_{2(p+1)+1} = &\sum_{k=1}^{2p+3}(-1)^{k-1}|b_k| \\
&=\sum_{k=1}^{2p+1}(-1)^{k-1}|b_k| \underbrace{ - |b_{2p+2}|+|b_{2p+3}|}_{\le0}\\
&\le S_{2p+1}
\end{align}
$$
Which means $S_{2p+1}$ is monotonically decreasing.
Now use the same approach for even number of terms:
$$
\begin{align}
S_{2(p+1)} = &\sum_{k=1}^{2p+2}(-1)^{k-1}|b_k| \\
&=\sum_{k=1}^{2p}(-1)^{k-1}|b_k| + \underbrace{ |b_{2p+1}|-|b_{2p+2}|}_{\ge0}\\
&\ge S_{2p}
\end{align}
$$
Which means $S_{2p}$ is monotonically increasing. Consider the following:
$$
S_{2p+1} - S_{2p} = |b_{2p+1}| \ge 0\tag{*}
$$
Now $S_2 = |b_1| - |b_2|$ and by monotoninity of $S_{2p}$:
$$
S_2 = |b_1| - |b_2| \le S_{2p}
$$
At the same time $S_1 = |b_1|$ which again by monotonicity of $S_{2p+1}$:
$$
S_{2p+1} \le S_1 = |b_1|
$$
Now using (*):
$$
|b_1| - |b_2| \le S_{2p} \le S_{2p+1} \le |b_1|\\
S_{2p-1} - S_{2p} = |b_{2p+1}| = |a_{2p+1} - a_{2p}|
$$
But $|a_{2p+1} - a_{2p}|$ is convergent as far as $(a_{2p+1} - a_{2p})$ is. Which yields:
$$
\lim_{p\to\infty}(S_{2p+1} - S_{2p}) = \lim_{p\to\infty}|b_{2p+1}| = \lim_{p\to\infty}|a_{2p+1} - a_{2p}| = 0
$$
Finally:
$$
\exists \lim_{n\to\infty}S_n \implies \exists \lim_{n\to\infty}\sum_{k=1}^n b_k = L
$$
Thus:
$$
\lim_{n\to\infty}\sum_{k=1}^n b_k = \lim_{n\to\infty}(a_n - a_1) = L \\
\implies \lim_{n\to\infty}a_n = L + a_1
$$
Thus $\{a_n\}$ is convergent.
