Proving a theorem about divergent sequences Suppose that $(a_n)$ and $(b_n)$ are increasing sequences of non-negative numbers and there is natural number $k$ such that
$$n\geq k \implies a_n \leq b_n$$
Prove that if $(b_n)$ is convergent then $(a_n)$ is convergent also. I tried to prove the contrary is impossible but I couldn't proceed.
 A: $\{a_n\}$ is an increasing sequence eventually bounded by the limit of $\{b_n\}$. Hence $\{a_n\}$ is convergent.
A: For $ n\ge k$, put
$$A_n=a_k+a_{k+1}+...+a_n$$
$$B_n=b_k+b_{k+1}+...+b_n$$
so
$$A_n\le B_n.$$
$a_n\ge 0 $ $\;\implies
  \; (A_n)_{n\ge k} $ is increasing.
$$\sum a_n \text{ diverges } \;\implies$$
$$\lim_{n\to +\infty}A_n=+\infty\; \implies$$
$$\lim_{n\to +\infty}B_n=+\infty\;\implies$$
$$\sum b_n \text{ diverges}$$
A: Fact 1. Let $l=\lim_{n\to\infty}b_n$. Then $b_n\le l$, for every $n$.
Moreover, for $m\le k$, we have $a_m\le a_k\le b_k$, and therefore $a_n\le l$, for every $n$.
So $(a_n)$ is an increasing and bounded sequence.
Fact 2. An increasing bounded sequence converges.

Both facts above are consequence of the following statement.
Theorem. If $(a_n)$ is an increasing and (upper) bounded sequence, then
$$
\lim_{n\to\infty} a_n=\sup\{a_n:n\in\mathbb{N}\}
$$
Proof. Surely the supremum exists, because the sequence is upper bounded. Call it $l$. Let $\varepsilon>0$. Then there exists $k$ such that $l-\varepsilon<a_k<l$; since the sequence is increasing, this implies $l-\varepsilon<a_n<l$, for every $n\ge k$. Hence $l=\lim_{n\to\infty}a_n$. QED
