# Proof verification: if $a_n, b_n>0$ and $\lim\limits_{n \to\infty} \frac{a_n}{b_n}=L_1$ with $L_1>0$, then if $\sum a_n$ converges, so does $\sum b_n$

I'm trying a proof technique I'm not used to for limits on fractions, which attempts to avoid an epsilon-delta approach similarly to how the single variable chain rule is proved in baby Rudin, and I was wondering if it works. Any help or tips are very welcome!

Statement: If $$a_n, b_n>0$$ and $$\lim \limits_{n \to \infty} \frac{a_n}{b_n} = L_1$$ with $$L_1 > 0$$, then if $$\sum_{n \in \mathbb N} a_n$$ converges, so does $$\sum_{n \in \Bbb N} b_n$$

Proof: Suppose $$\sum_{n \in \Bbb N} a_n$$ converges to $$L_a$$.

Since $$\lim \limits_{n \to \infty} \frac{a_n}{b_n} = L_1$$, we have $$a_n = b_n(L_1 + \varepsilon(n))$$ with $$\varepsilon(n) \to 0$$ as $$n \to \infty$$.

Therefore, $$\sum_{n \in \Bbb N} a_n = \sum_{n \in \Bbb N} b_n(L_1 + \varepsilon(n)) = L_a,$$

so $$\lim \limits_{n \to \infty} \sum_{i=1}^n b_i(L_1 + \varepsilon(n)) = L_a,$$

and therefore,

$$\sum_{i=1}^nb_i=\frac{L_a+\mu(n)}{L_1 + \varepsilon(n) } ,$$ where $$\mu(n) \to 0$$ as $$n \to \infty$$.

Therefore, letting $$n \to \infty$$ results in $$\sum_{i=1}^nb_i = \frac{L_a}{L_1}$$, so $$\sum_{n \in \mathbb N}b_n$$ converges. $$\blacksquare$$

• Your proof appears to fail if $L_1 = 0$. Commented Oct 20, 2021 at 5:40
• @ClementYung ah, thanks for catching that! $k$ was a placeholder, i forgot to change it to $L_1$. It's updated now. I've also updated the premise, thanks for the comment! Commented Oct 20, 2021 at 5:41
• if $a_n \neq 0$, then write $b_n = \frac{b_n}{a_n} \times a_n$ and proceed Commented Oct 20, 2021 at 5:44
• It seems like you treated $L_1+\varepsilon(n)$ as a constant (you factored it out of the partial sums) after you deduced that $\sum b_n(L_1+\varepsilon(n))=L_a$. This is not justified in your argument. Moreover, your summation index changed from $n$ to $i$. Commented Oct 20, 2021 at 5:45
• $\lim \limits_{n \to \infty} \sum_{i=1}^n b_n(L_1 + \varepsilon(n)) = L_a$ does not imply $\sum_{i=1}^nb_n=\frac{L_a+\mu(n)}{L_1 + \varepsilon(n) }$ Commented Oct 20, 2021 at 5:48

$$\sum_{n \in \Bbb N} a_n = \sum_{n \in \Bbb N} b_n(L_1 + \varepsilon(n)) = L_a$$, so $$\lim \limits_{n \to \infty} \sum_{i=1}^n b_i(L_1 + \varepsilon(n)) = L_a$$

Actually, you should have $$\lim_{n\to\infty}\sum_{i=1}^n b_i(L_1+\epsilon(\color{red}{i}))=L_a$$

(note the red colored $$i$$ inside the sum).

Also, even if the continuation would be correct, you are jumping over a lot of steps.

$$\lim \limits_{n \to \infty} \sum_{i=1}^n b_i(L_1 + \varepsilon(n)) = L_a$$, and therefore, $$\sum_{i=1}^nb_i=\frac{L_a+\mu(n)}{L_1 + \varepsilon(n) }$$, where $$\mu(n) \to 0$$ as $$n \to \infty$$

This is a big leap. I don't see where you get the latter expression from the former. You really should expand this out to explain exactly how the limit implies that the sum is equal to $$\frac{L_a+\mu(n)}{L_1 + \varepsilon(n)}$$, and explain exactly what $$\mu(n)$$ is.

• I think the OPs inadvertent summation index change led them to conclude that $L_1+\varepsilon(n)$ is constant in the eyes of the summation. Commented Oct 20, 2021 at 6:00
• @AlannRosas Probably, yeah. But I still wanted to point out that skipping over too many steps is not ok when writing proofs.
– 5xum
Commented Oct 20, 2021 at 6:01
• @Koro changing the name of the variables, we get $a_i = b_i(L_1 + \epsilon(i))$ with $\epsilon(i) \to 0$ as $i \to \infty$, true for any $i$ or $\epsilon(i)$. Only then can we produce the summation: $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i(L_1+\epsilon(i))$, which is always true no matter the size of $\epsilon(i)$ or $i$ or $n$, but my mistake was to conclude instead $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i(L_1+\epsilon(n))$ Commented Oct 20, 2021 at 6:10
• @Koro No, we can say $a_1=b_1(L_1+\varepsilon(1))$. By definition, this is true, because by definition, $\varepsilon(n) = \frac{a_n}{b_n} - L_1$. For large values of $n$, it is also true that $\varepsilon(n)$ is small, (and this is not necessarily true for small values of $n$), but the equation holds for all $n$.
– 5xum
Commented Oct 20, 2021 at 6:11
• @Koro Exactly, $\varepsilon(n)$ is simply some sequence that converges to $0$. In fact, that is all we know of $\varepsilon(n)$.
– 5xum
Commented Oct 20, 2021 at 6:12