$\lim\limits_{n \to \infty} \frac{a_n}{b_n}=l$ ; Prove $\sum\limits_{k=1}^\infty a_k$ converges iff $\sum\limits_{k=1}^\infty b_k$ converges.

Solution 1:

As you wrote, both sequences $\dfrac{a_n}{b_n}, \dfrac{b_n}{a_n} $ converge. And convergent sequences are bounded. Thus both sequences $\dfrac{a_n}{b_n}, \dfrac{b_n}{a_n} $ are bounded, say by $A$ and $B$ respectively. This allows us to prove the result quickly: Suppose $\sum a_n$ converges. Since

$$b_n = \dfrac{b_n}{a_n}a_n \le Ba_n\,\,\text { for all }n,$$

we have for any $N,$

$$\sum_{n=1}^{N} b_n \le \sum_{n=1}^{N} Ba_n = B\sum_{n=1}^{N} a_n <B\sum_{n=1}^{\infty} a_n < \infty.$$

Thus the sequence of partial sums of $\sum b_n,$ which are monotonically increasing, are bounded above. Hence this sequence converges. By definition, that means $\sum_{n=1}^{\infty} b_n$ converges. Of course the proof the other way around is the same, and we're done.

Solution 2:

Overall looks ok. But you can simplify the proof a lot just from the fact that $$\left|\frac{a_k}{b_k}-l\right|<\varepsilon \Leftrightarrow (l-\varepsilon)b_k<a_k<(l+\varepsilon)b_k, \forall k>N(\varepsilon)$$ 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 $l-\varepsilon>0$.