can someone verify following proof of abel's theorem
Abel's convergence theorem : If $\sum_{i=1}^{\infty}a_n$ converges and $(b_n)$ is monotonic convergent sequence then $\sum_{i=1}^{\infty} a_n b_n$ converges
we know that by summation by parts we have that $\sum_{j=m+1}^n a_j b_j = A_n b_{n+1} - A_mb_{m+1} + \sum_{j=m+1}^n A_j (b_i - b_{i+1})$ where $A_n = \sum_{i=1}^n a_i$
we will prove this by cauchy crtierion for convergence of series
so we have to show that $\forall \epsilon>0 \,\, \exists N \,\, \in \mathbb{N} \,\, \forall n > m \geq N \,\,|\sum_{j=m+1}^n a_jb_j| < \epsilon$
$= |\sum_{j=m+1}^n a_jb_j|$
$ \leq |A_n b_{n+1} - A_m b_{m+1}| + \sum_{j=m+1}^n |A_j||b_i - b_{i+1}|$
key point is that $(A_nb_{n+1})$ is convergent sequence cause it is product of two convergent sequences
now by cauchy criterion for convergence of sequences we have that there exists $N_1 \in \mathbb{N} \,\, \forall n > m \geq N |A_{n}b_{n+1} - A_{m}b_{m+1}| < \epsilon/2$
now for other term $\sum_{j=m+1}^n |A_j||b_i - b_{i+1}| \leq M \sum_{j=m+1}^n |b_i - b_{i+1}| = |b_{n+1} - b_{m+1}|$ where $M$ is upper bound for partial sums of $\sum_{n}a_n$ and sum simplifies due to montonicity
which we can apply cauchy criterion for convergence of sequences we have $\exists N_2 \in \mathbb{N} \,\, \forall n>m\geq N \,\, |b_{n} - b_{m}| < \frac{\epsilon}{2M}$
- our required $N = \max\{N_1, N_2\}$
Is there any flaw in above proof ? thanks in advance