in Problems in Mathematical Analysis I problem 2.3.16 a),
if $\lim\limits_{n \to \infty}a_n =a$, then find $\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}$
The proof that was on the answers:- by Toeplitz theorem let $c_{n,k} =\frac{1}{(n+1-k)(n+2-k)}$ and the answer is $a$, here I didn't understand how this is possible because one condition for Toeplitz theorem is $ \lim\limits_{n \to \infty} c_{n,k} =0 \ \forall k \in \mathbb{N}$ doesn't hold if $k=n $ for example
$\textbf{Toeplitz theorem :}$ let $\{ c_{n,k}: 1\leq k \leq n \}$ be an array of real numbers such that:
1-$ \lim\limits_{n \to \infty} c_{n,k} =0 \ \forall k \in \mathbb{N}$
2- $\sum\limits_{k=1} ^{n} c_{n,k} \to 1$ as $n\to \infty$
3- There is exist $C \in \mathbb{R}$ such that for all positive integer $n$ $\sum\limits_{k=1} ^{n} |c_{n,k}| \leq C$
then for any converging sequence $ \{a_n \}$ the sequence $\{ b_n \}$ given by $b_n:=\sum\limits_{k=1} ^{n} c_{n,k} a_k $ is convergent and $\lim\limits_{n \to \infty}b_n= \lim\limits_{n \to \infty}a_n$
$\textbf{ My Proof:-}$
First I have to prove that the sequence $\{ b_n\}$ is convergent this can be shown by considering the sum $s_n =\sum\limits_{k=1} ^{n} |c_{n,k} a_k| $ since the terms of the sum consist only positive number and since $a_n$ is convergent sequence then there exist some $M \in \mathbb{R} $ such that $a_k\leq M \ \forall k $ then the sequence $s_n <M , \sum\limits_{k=1} ^{n} |c_{n,k}| \leq C M $ then the sequence $s_n $ converge then $b_n$ converge Now lets proof that $\lim\limits_{n \to \infty}b_n= \lim\limits_{n \to \infty}a_n$ first let $a:=\lim\limits_{n \to \infty}a_n$ since $\sum\limits_{k=1} ^{n} c_{n,k} \to 1$ as $n\to \infty $ then I need to prove that $ \sum\limits_{k=1} ^{n} c_{n,k} a_k-a \sum\limits_{k=1} ^{n} c_{n,k} \to 0 \text{ as } n \to \infty $ $$S_n:=\sum\limits_{k=1} ^{n} c_{n,k} (a_k-a)$$ $$\text{since } a_k \to a \text{ then } \forall \varepsilon_1 >0 \ \exists N \in \mathbb{N} \text{ such that } \forall n \geq N \ |a_k-a|<\varepsilon_1 $$ $$ \text{Choose $\varepsilon =\inf \{ \varepsilon_1 , \frac{\varepsilon_1}{ C M}\}$ } \text{Choose $n$ such that } \sup\{ |c_{n,k}| \} < \frac{\varepsilon_1}{NM}$$ $$-\varepsilon - \sum\limits_{k=N+1} ^{n} |c_{n,k} a_k| \leq S_n \leq \varepsilon + \sum\limits_{k=N+1} ^{n} |c_{n,k} a_k|$$ since $\sum\limits_{k=N+1} ^{n} |c_{n,k} a_k| <\varepsilon$ then $$-2\varepsilon<S_n< 2\varepsilon$$
I tried to prove it myself and this what I got
let $b_{n-1}= a_n - a_{n-1}$ it is easy to see that $\lim\limits_{n \to \infty }b_n =0 $ (because any converging sequence in $\mathbb{R}$ is a Cauchy sequence) so $$ \lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}=\lim\limits_{n \to \infty} \left( a_n - \frac{a_1}{n+1} +b_n -\sum\limits_{k =1} ^n \frac{b_k}{n+1-k} \right)=a -\lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k} $$ and here I couldn't prove that $\lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k} =0$
I also want to ask why is the answers that the book provide is correct ? because one of the necessary condition of Toeplitz theorem doesn't hold
EDIT
@TheSilverDoe has an excellent answer but now I don't know where my mistake was since $\lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k} \neq 0$ and $$ \lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}= \lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)}- \frac{a_k}{(n+2-k)}$$ $$ =\lim\limits_{n \to \infty}+a_n -\frac{a_1}{n+1}- \sum\limits_{k=1 }^{n-1} \frac{a_{n+1-k}- a_{n-k}}{k+1} $$ $$= a-\lim\limits_{n \to \infty} \sum_{k=1}^{n-1}\frac{a_{k+1}- a_{k}}{n+1-k}=a-\lim\limits_{n \to \infty} \sum_{k=1}^{n-1}\frac{b_{k}}{n+1-k}=a-\lim\limits_{n \to \infty} \sum_{k=1}^{n}\frac{b_{k}}{n+1-k}- b_{n}=$$ $$a- \lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}$$
I still can't figure out where is the mistake in my logic.
Is the mistake related to conditional convergent series and Riemann rearrangement theorem ?