How can I prove $|a_{m+1}-a_{m+2}+a_{m+3}-...\pm a_{n}| \le |a_{m+1}|$? Give a sequence $(a_n)$ that satisfying:
(i) the sequence is decreasing: $$a_n \ge a_{n+1} \forall n \in \mathbb N$$
(ii) the sequence is converging to $0$: $$(a_n) \rightarrow 0$$
Problem:Prove that $|a_{m+1}-a_{m+2}+a_{m+3}-...\pm a_{n}| \le |a_{m+1}|$ $\forall n \gt m \ge N$ for some $N \in \mathbb N$
I think we shall do this by induction. Let $n=1$, then clearly $|a_{m+1}| \le |a_{m+1}|$. Suppose $n=k$ holds true, then for $n=k+1$,
$|a_{m+1}-a_{m+2}+a_{m+3}-...\pm a_{m+k} \mp a_{m+k+1}|$ ...
For the step $n=k+1$, I give up(the triangle inequality will not work!). So, how do we continue to do this? I have a strange feeling that I have made a big mistake somewhere because this induction is really weird. Please help me proceed with how to prove this strange inequality. I thank you very much.
Note: every term in this sequence MUST BE positive(proven). 
 A: Let us write $n=m+p$, so that the sum has $p$ elements. We prove that 
$$\left| a_{m+1}-a_{m+2} +\cdots \pm  a_{m+p} \right| \leq |a_{m+1}|=a_{m+1} \tag{1} $$ for all $m,p$:
First note that the $\pm$ actually equals $(+)$ if $p$ is odd, and is $(-)$ if $p$ is even. This motivates breaking the proof into two cases:


*

*If $p=2k+1$ is odd, summing the inequalities $a_{m+2l+1}-a_{m+2l+2} \geq 0$ from $l=0$ to $k-1$, shows that we can take the absolute value in $(1)$ off, as both expressions
$$(a_{m+1}-a_{m+2})+(a_{m+3}-a_{m+4})+ \cdots (a_{m+p-2}-a_{m+p-1}) $$
and
$$a_{m+p} $$
are nonnegative in that case. Once the absolute value is gone, it's easy to see that $(1)$ holds by considering both expressions above again.

*If $p=2k$ is even, we can similarly break the sum into a sum of nonnegative summands, which allows us to take the absolute value off. Can you take it from here?
A: This is just an induction form of the previous answer:

We will show that $0\le a_{m+1}-a_{m+2}+a_{m+3}-\cdots+(-1)^{k-1}a_{m+k}\le a_{m+1}$
for all integers $k\ge1 \text{ and } m\ge0$ by induction on $k$:
1) This is valid for $k=1$, since $0\le a_{m+1}\le a_{m+1}$.
2) Assume this is valid for all integers $l$ with $1\le l\le k$.
a)  If k is even, then 
i) $\displaystyle\sum_{i=1}^{k+1}(-1)^{i-1}a_{m+i}=\sum_{i=1}^{k}(-1)^{i-1}a_{m+i}+a_{m+k+1}\ge0$ and
ii) $\displaystyle\sum_{i=1}^{k+1}(-1)^{i-1}a_{m+i}=a_{m+1}-\sum_{i=1}^{k}(-1)^{i-1}a_{m+i}\le a_{m+1}$  $\;$since $\displaystyle\sum_{i=1}^{k}(-1)^{i-1}a_{m+i}\ge 0$
b)  If k is odd, then
i) $\displaystyle\sum_{i=1}^{k+1}(-1)^{i-1}a_{m+i}=\sum_{i=1}^{k-1}(-1)^{i-1}a_{m+i}+(a_{m+k}-a_{m+k+1})\ge0$ and
ii) $\displaystyle\sum_{i=1}^{k+1}(-1)^{i-1}a_{m+i}=\sum_{i=1}^{k}(-1)^{i-1}a_{m+i}-a_{m+k+1}\le \sum_{i=1}^{k}(-1)^{i-1}a_{m+i}\le a_{m+1}$
