Fdtxjr

Let $a_n, ninBbb N$ be a sequence such that:
$$
lim_ntoinfty(a_n+1 - a_n) = 0tag1
$$
The sequence also satisfies the following properties:
$$
beginalign*
|a_n+2-a_n+1| le |a_n+1-a_n|tag 2\
(a_n+2 - a_n+1)(a_n+1-a_n) le 0 tag 3
endalign*
$$
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:
$$
a_n = sum_k=1^n1over k
$$
Which is clearly divergent as $ntoinfty$, 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:
$$
b_n = a_n - a_n-1tag 4
$$
If a sequence converges then its absolute values also converge:
$$
lim_ntoinftyb_n = 0 implies lim_ntoinfty|b_n| = 0
$$
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:
$$
b_n+2b_n+1 le 0\
b_n+1b_n le 0\
b_nb_n-1 le 0\
cdots
$$
Let's stick to case $(5)$ for definiteness. This implies:
$$
0 le b_n+2le b_n le b_n-2 le cdots tag7
$$
At the same time since $|b_n|$ is monotically decreasing and $b_n+1le 0$:
$$
cdots le b_n-3 le b_n-1 le b_n+1 cdots le 0 tag8
$$

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?

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.

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:
$$
lim_ntoinftyb_n = 0\
|b_n+2| le |b_n+1|\
b_n+2b_n+1 le 0
$$
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:
$$
beginalign*
S_2p+1 = sum_k=1^2p+1(-1)^k-1|b_k| \
S_2p = sum_k=1^2p(-1)^k-1|b_k|
endalign*
$$

Consider the sum of odd amount of terms:
$$
beginalign
S_2(p+1)+1 = &sum_k=1^2p+3(-1)^k-1|b_k| \
&=sum_k=1^2p+1(-1)^k-1|b_k| underbrace_le0\
&le S_2p+1
endalign
$$
Which means $S_2p+1$ is monotonically decreasing.

Now use the same approach for even number of terms:
$$
beginalign
S_2(p+1) = &sum_k=1^2p+2(-1)^k-1|b_k| \
&=sum_k=1^2p(-1)^k-1|b_k| + underbraceb_2p+2_ge0\
&ge S_2p
endalign
$$
Which means $S_2p$ is monotonically increasing. Consider the following:
$$
S_2p+1 - S_2p = |b_2p+1| ge 0tag*
$$

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:
$$
lim_ptoinfty(S_2p+1 - S_2p) = lim_ptoinfty|b_2p+1| = lim_ptoinfty|a_2p+1 - a_2p| = 0
$$

Finally:
$$
exists lim_ntoinftyS_n implies exists lim_ntoinftysum_k=1^n b_k = L
$$
Thus:
$$
lim_ntoinftysum_k=1^n b_k = lim_ntoinfty(a_n - a_1) = L \
implies lim_ntoinftya_n = L + a_1
$$

Thus $a_n$ is convergent.