Is there a difference between these two statements concerning sequences Suppose we have a sequence of positive real numbers $\{a_n\}$ and $\{b_n\}$. 
I want to compare the following statements:
\begin{align}
\forall \, m\in\mathbb{N}: b_m &\leq b_{m+1} \leq a_{m+1} \leq a_m \hspace{10mm} (1)\\[5pt]
\forall \, m,n \in \mathbb{N}: b_m &\leq b_{m+1} \leq a_{n+1} \leq a_n \hspace{12mm} (2)
\end{align}
Is my assessment written below correct?
To me, $(1)$ is saying $\{b_m\}$ is at most equal to the corresponding term, and every term thereafter, in $\{a_m\}$. However, I think $(1)$ still leaves open the possibility that distinct indices $i$ and $j$ exist such that $a_i < b_j$. Is that correct?
In contrast $(2)$ says that every terms in $\{b_n\}$ is always at most equal to every term in $\{a_n\}$----no exceptions. 
 A: I'd say that there are two questions here: (a) What do these statements say immediately; (b) What can be deduced from them?
If we just look at what they say, there are different. They do have a lot in common: the "$b_m\le b_{m+1}$" part says that $\{b_n\}$ is a non-decreasing sequence; and the "$a_{m+1}\le a_m$" or "$a_{n+1}\le a_n$" part says that $\{a_n\}$ is a non-increasing sequence. The difference is in the middle inequality only, and your interpretation is almost correct. More specifically:

To me, (1) is saying $\{b_m\}$ is at most equal to the corresponding term … in $\{a_m\}$. However, I think (1) still leaves open the possibility that distinct indices $i$ and $j$ exist such that $a_i<b_j$.
In contrast (2) says that every terms in $\{b_n\}$ is always at most equal to every term in $\{a_n\}$ — no exceptions.

Yes, that's absolutely correct! But:

… and every term thereafter …

is not exactly right. If you logically continue indexing:
$$b_{m+1}\le a_{m+1}\le a_m\le a_{m-1}\le a_{m-2}\le\cdots,$$
you will see that $b_m$ is at most equal to the corresponding term $a_m$ and all preceding terms, not further terms, of the sequence $\{a_n\}$.
That being said, i.e. even though directly these two statement say different things, in fact with some reasoning we can deduce that they are equivalent to each other. That (2) implies (1) is obvious, simply because (2) is more general than (1). But we can also deduce that (1) implies (2), i.e. that $b_i\le a_j$ for all $i,j$:

*

*If $i>j$, then setting $i=m+1$, we can deduce that $b_i\le a_i\le a_{i-1}\le a_{i-2}\le\cdots\le a_j$;

*If $i<j$, then setting $j=m+1$, we can deduce that $b_i\le b_{i+1}\le b_{i+2}\le\cdots\le b_j\le a_j$.

