Proving Abel's test of convergence I am self-learning Real Analysis from the text, Understanding Analysis, by Stephen Abbott. I would like to prove Abel's test of convergence in the following exercise problem 2.7.12. A clue/hint, without revealing the entire solution would be extremely helpful.
I reproduce my attempt for part(a) of the problem below.

(Abel's Test). Abel's test for convergence states that if the series $\displaystyle \sum _{k=1}^{\infty } x_{k}$ converges and if $\displaystyle ( y_{k})$ is a sequence satisfying:


\begin{equation*}
y_{1} \geq y_{2} \geq y_{3} \geq \dotsc \geq 0
\end{equation*}
then the series $\displaystyle \sum _{k=1}^{\infty } x_{k} y_{k}$ converges.
(a) Use exercise 2.7.12 to show that:


\begin{equation*}
\sum _{k=1}^{n} x_{k} y_{k} =s_{n} y_{n+1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1})
\end{equation*}

where $\displaystyle s_{n} =x_{1} +\dotsc +x_{n}$.
Proof.
By the formula for summation by parts, we have:
\begin{equation*}
\begin{array}{ c l }
\sum _{k=1}^{n} x_{k} y_{k} & =s_{n} y_{n+1} -s_{0} y_{1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1})\\
 & =s_{n} y_{n+1} -0\cdot y_{1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1})\\
 & =s_{n} y_{n+1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1})
\end{array}
\end{equation*}

(b) Use the comparison test to argue that: $\displaystyle \sum _{k=1}^{\infty } s_{k}( y_{k} -y_{k+1})$ converges absolutely, and show how this leads directly to a proof of Abel's test.

 A: My proof attempt for the sake of completeness.
Proof.
Since $\displaystyle \sum _{k=1}^{\infty } x_{k}$ converges, $\displaystyle ( s_{k})$ is a convergent sequence and hence it is bounded. There exists $\displaystyle M >0$ for all $\displaystyle k\in \mathbf{N}$, such that $\displaystyle |s_{k} |\leq M$. Thus,
\begin{equation*}
|s_{k}( y_{k} -y_{k+1}) |\leq M|y_{k} -y_{k+1} |\leq M( y_{k} -y_{k+1}) \quad \{\because y_{k} \geq y_{k+1}\}
\end{equation*}
We know that $\displaystyle ( y_{k} -y_{k+1}) \geq 0$. Let $\displaystyle ( t_{k})$ be the sequence of partial sums of the infinite series $\displaystyle \sum _{k=1}^{\infty }( y_{k} -y_{k+1})$. Since $\displaystyle ( t_{k})$ is a monotonically-increasing sequence and
\begin{equation*}
\begin{array}{ c l }
t_{k} & =( y_{1} -y_{2}) +( y_{2} -y_{3}) +\dotsc +( y_{k} -y_{k+1})\\
 & =y_{1} -y_{2} +y_{2} -y_{3} +y_{3} +\dotsc +y_{k} -y_{k+1}\\
 & =y_{1} -y_{k+1}\\
 & \leq y_{1}
\end{array}
\end{equation*}
it is bounded by $\displaystyle y_{1}$, by the Montone convergence theorem, $\displaystyle t_{k}$ is a convergent series. By the Algebraic limit theorem for infinite series, $\displaystyle M\cdot \sum _{k=1}^{\infty }( y_{k} -y_{k+1})$ is a convergent series.
Hence, by the comparison test, $\displaystyle \sum _{k=1}^{\infty } |s_{k}( y_{k} -y_{k+1}) |$ is a convergent series.
By the Absolute convergence test, $\displaystyle \sum _{k=1}^{\infty } s_{k}( y_{k} -y_{k+1})$ is a convergent series.
Passing to the limits, we have:
\begin{equation*}
\begin{array}{ c c }
\lim _{n\rightarrow \infty }\sum _{k=1}^{n} x_{k} y_{k} & =\lim _{n\rightarrow \infty }\left[ s_{n} y_{n+1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1})\right]
\end{array}
\end{equation*}
Note that, $\displaystyle ( y_{n})$ is bounded by $\displaystyle [ 0,y_{1}]$ and is a monotonically decreasing sequence. Hence, by MCT, it is a convergent sequence. $\displaystyle ( s_{n})$ is also given to be a convergent sequence. Hence, the limit of the right hand side of the expression can be written as:
\begin{equation*}
\lim _{n\rightarrow \infty }\left[ s_{n} y_{n+1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1})\right] =\lim _{n\rightarrow \infty } s_{n} y_{n+1} +\lim _{n\rightarrow \infty }\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1})
\end{equation*}
Since both these limits exist, the limit on the right hand side exists. Thus, the product series $\displaystyle \sum _{k=1}^{\infty } x_{k} y_{k}$ converges.
