Why Is the Lower Bound of the Cauchy Product Greater Than w_{⌊n/2⌋}?

[Bleys]

I don't understand a small part in the proof that two absolutely convergent series have absolutely convergent cauchy product.
Instead of writing the whole thing, I'll write the essentials and the step I'm having trouble with.

\sum_{r=1}^{\infty}a_{r} and \sum_{r=1}^{\infty}b_{r} are positive term series that are absolutely convergent. Denote their partial sums as s_{n} , t_{n} respectively. Let w_{n}=s_{n}t_{n} and u_{n}=\sum_{r=1}^{\n}c_{r} where c_{n} is the Cauchy product of a_{n} and b_{n}

Then w_{\lfloor n/2\rfloor}\leq u_{n}\leq w_{n}. This is the step I don't understand. I can see why it would be smaller than w_{n}, since it's a sum containing u_{n}, but I don't see why it would be greater than w_{\lfloor n/2\rfloor}.

 

[fresh_42]

Mertens' Theorem

Let's consider two absolute convergent series $A,B$, although only the absolute convergence of $A$ is needed. Let's note the partial sums $A_n=\sum_{k=0}^n a_k\, , \,B_n=\sum_{k=0}^n b_k$.
\begin{align*}
AB&=(A-A_n)B +\sum_{k=0}^n a_kB\\
S_n&=\sum_{k=0}^n c_k = \sum_{k=0}^n \sum_{j=0}^ka_jb_{k-j}=\sum_{k=0}^n a_kB_{n-k} \\
AB-S_n&= (A-A_n)B+\sum_{k=0}^n a_k(B-B_{n-k})
\end{align*}
The first term converges to $0$ and with $N:=\lfloor \dfrac{n}{2} \rfloor$ we can write the second term
$$
\sum_{k=0}^N (B-B_{n-k}) = \underbrace{\sum_{k=0}^n a_k(B-B_{n-k})}_{=P_n}+\underbrace{\sum_{k=N+1}^n a_k(B-B_{n-k})}_{=Q_n}
$$
For $P_n$ we have
$$
|P_n| \leq \sum_{k=0}^N|a_k|\cdot |B-B_{n-k}|\leq \max_{N\leq k \leq n}|B-B_k|\cdot \sum_{k=0}^N|a_k| \longrightarrow 0
$$
because $A$ converges absolutely and $(B-B_k)_k$ is a bounded sequence converging to $0$, i.e. there is a constant $c$ such that $|B-B_k|<c$ for all $k\in \mathbb{N}_0\,.$ Therefore we get
$$
|Q_n|\leq \sum_{k=N+1}^n |a_k|\cdot |B-B_{n-k}| \leq c \sum_{k=N+1}^n |a_k| \longrightarrow 0
$$
by the Cauchy criterion. Hence $AB-S_n \longrightarrow 0$ or $S_n \longrightarrow AB\,.$