real analysis - $limlimits_{n to infty} frac{a_n}{b_n}=l$ ; Prove $sumlimits_{k=1}^infty a_k$ converges iff $sumlimits_{k=1}^infty b_k$ converges.

Written, in Ex.8 Ch.9.1 of the book Advanced Calculus by P. M. Fitzpatrick :

Suppose that $\sum\limits_{k=1}^\infty a_k$ and $\sum\limits_{k=1}^\infty b_k$ are series of positive numbers such that $$\lim_{n \to \infty} \frac{a_n}{b_n}=l \ \ \ \text{and} \ l>0.$$ Prove that the series $\sum\limits_{k=1}^\infty a_k$ converges iff the series $\sum\limits_{k=1}^\infty b_k$ converges.

Am I correct by the following (sketch of) proof? :

1- For a given $\epsilon_1$ there is $N_1$ such that $\left|\frac{a_n}{b_n} - l\right| < \epsilon_1$ for all $n \ge N_1$.

2- Since the series $\sum\limits_{k=1}^\infty a_k$ converges, for a given $\epsilon_2$ there is $N_2$ such that $\left|b_{n+1}+\dots+b_{n+k}\right|< \epsilon_1$ for all $n \ge N_2$ any for all natural numbers $k$.

3- Define $N = \max {\{N_1,N_2}\}$.

4- From Step (1), $a_{n+k} < (\epsilon_1+l) b_{n+k}$ for all $n \ge N$ any for all natural numbers $k$. [Also, $a_i$'s and $b_i$'s are all positive]. Thus $a_{n+1}+\dots+a_{n+k} < (\epsilon_1+l) (b_{n+1}+\dots+b_{n+k})< \epsilon_3$. Then the convergence of the series $\sum\limits_{k=1}^\infty b_k$ implies the convergence of the series $\sum\limits_{k=1}^\infty a_k$.

5- For the reverse implication, we use the fact that $\lim\limits_{n \to \infty}\frac{a_n}{b_n}=l \iff \lim\limits_{n \to \infty}\frac{b_n}{a_n}= \frac{1}{l}= l' >0$ and repeat the process this time for a and b exchanged.

6- The Quotient Property For Sequences hold for a nonzero limit in the denominator, but since the limit in the numerator also is zero so we may use the The Quotient Property For Sequences.

Thanks.

Answer

Overall looks ok. But you can simplify the proof a lot just from the fact that
$$\left|\frac{a_k}{b_k}-l\right|<\varepsilon \Leftrightarrow (l-\varepsilon)b_kN(\varepsilon)$$
Which means:

1. If $\sum_{k=1}^{\infty}b_k<\infty$ then
   $$\sum_{k=1}^{\infty}a_k=\sum_{k=1}^{N(\varepsilon)}a_k+\sum_{k=N(\varepsilon)+1}^{\infty}a_k<\sum_{k=1}^{N(\varepsilon)}a_k + (l+\varepsilon)\sum_{k=N(\varepsilon)+1}^{\infty}b_k<\infty$$




2. If $\sum_{k=1}^{\infty}a_k<\infty$ then
   $$\sum_{k=1}^{\infty}b_k=\sum_{k=1}^{N(\varepsilon)}b_k+\sum_{k=N(\varepsilon)+1}^{\infty}b_k<\sum_{k=1}^{N(\varepsilon)}b_k + \frac{1}{l-\varepsilon}\sum_{k=N(\varepsilon)+1}^{\infty}a_k<\infty$$
   $\varepsilon>0$ can be small enough so that $l-\varepsilon>0$.