If $\sum a_n$ is divergent , $b_n \uparrow \infty $ then $\Sigma a_n b_n $ is divergent?

If $\sum a_n$ is divergent, $b_n$ is unbounded and increasing then $\sum a_n b_n $ is divergent?

I think it is true.

Given $\sum a_n$ is divergent, so the sequence of partial sums $S_n=\sum_{k=1}^n a_k $ is divergent.

Thus $\exists \epsilon > 0$ such that $\forall k\in \mathbb{N} , \exists m_k , n_k $ such that $| S_{n_k} - S_{m_k} |> \epsilon$.

Let $T_n=\sum_{k=1}^n a_k b_k =\sum_{k=1}^n (S_{k+1} - S_k)b_k$.

Let $k$ be large enough such that $n_k > m_k \implies b_{n_k} > b_{m_k}>0$.

Now I want to show $| T_{n_k} - T_{m_k} | > \epsilon \times \lambda $, where $\lambda $ is some constant but unable to proceed with the telescoping nature of $S_n$.

I am mainly facing problem with the sign of $a_n$.

Or is the assertion false? Any hints would be appreciated.