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). 

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) $\displaystyle\sum_{i=1}^{k+1}(-1)^{i-1}a_{m+i}=\sum_{i=1}^{k}(-1)^{i-1}a_{m+i}-a_{m+k+1}\le \sum_{i=1}^{k}(-1)^{i-1}a_{m+i}\le a_{m+1}$
