2.7.14 Abbott Abel's Convergence Test proof 
Abel's test: $\sum x_n $ converges and $(y_n)$ is a monotonically decreasing positive sequence. Then $\sum x_ny_n $ converges. 

$\sum x_n $ is bounded by some $B$ (as it converges) and $y_n$ converges to something $\ge 0$. Using Summation by parts (2.7.12) we can show $|\sum_{j=m+1} ^n x_j y_j| \le 2 B y_{1} $.
Now we know that  $\sum x_n$  converges so we can apply Cauchy to find $|\sum_{j=m+1}^{n} x_j| < \frac{\epsilon}{2y_1} $. The leap is to assume  $B\le |\sum_{j=m+1}^{n} x_j| < \frac{\epsilon}{2y_1}$. 
$\sum x_n $ is bounded by  $B$ i.e. $\sum x_n \le B$. How do you go from here to $B\le |\sum_{j=m+1}^{n} x_j|$?
Hints would be helpful.
 A: Some of your observations are correct and will be of use, but your estimate from partial summation is not helpful in proving convergence. You are on the right track trying to establish the Cauchy criterion, that for any $\epsilon > 0$ there is a positive integer $N$ such that for all $n> m> N$ we have
$$\left|\sum_{j=m+1}^n x_j y_j \right| < \epsilon.$$
Summing by parts with $S_n = \sum_{j=1}^n x_j$,we get
$$\tag{1}\sum_{j=m+1}^n x_j y_j = y_{n+1} S_n - y_{m+1} S_m + \sum_{j=m+1}^n S_j(y_j - y_{j+1}).$$
By hypothesis the series $\sum_j x_j$ converges to some number $S$ and there must be a bound $B> 0$ such that $\left| S_n \right| \leqslant B$ for all $n$.  Also we have that the sequence $(y_j)$ is positive and monotonically decreasing.  Hence, we have $y_j \leqslant y_1$ for all $j$ and convergence $y_j \to y \geqslant 0$ as $j \to \infty$ as you observed.
From (1) we get
$$\sum_{j=m+1}^n x_j y_j = S_n(y_{n+1} - y_{m+1}) - y_{m+1} (S_m - S_n) + \sum_{j=m+1}^n S_j(y_j - y_{j+1}),$$
Using the triangle inequality,
$$\tag{2}\begin{align} \left|\sum_{j=m+1}^n x_j y_j\right| &\leqslant  |S_n|\, |y_{n+1} - y_{m+1}| + y_{m+1} \, |S_m - S_n| + \sum_{j=m+1}^n |S_j| \, (y_j - y_{j+1})\\ &\leqslant B\, |y_{n+1} - y_{m+1}| + y_1 \, |S_m - S_n| + B\sum_{j=m+1}^n (y_j - y_{j+1}) \\ &=   B\, |y_{n+1} - y_{m+1}| + y_1 \, |S_m - S_n| + B(y_{m+1} - y_{n+1}) \\ &= y_1 \, |S_m - S_n| + 2B | y_{m+1} - y_{n+1}|\end{align}.$$
You should be able to finish now by using the fact that the sequences $(y_j)$ and $(S_n)$ are convergent, hence Cauchy, and show that there exists $N$ such that the right-hand side of (2) can be made smaller than $\epsilon$ for all $n > m > N.$
