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. $\blacksquare$

  • $\begingroup$ Your proof appears to fail if $L_1 = 0$. $\endgroup$ Commented Oct 20, 2021 at 5:40
  • $\begingroup$ @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! $\endgroup$
    – shintuku
    Commented Oct 20, 2021 at 5:41
  • $\begingroup$ if $a_n \neq 0$, then write $b_n = \frac{b_n}{a_n} \times a_n$ and proceed $\endgroup$
    – Math-fun
    Commented Oct 20, 2021 at 5:44
  • $\begingroup$ 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 $i$. $\endgroup$ Commented Oct 20, 2021 at 5:45
  • 3
    $\begingroup$ $\lim \limits_{n \to \infty} \sum_{i=1}^n b_n(L_1 + \varepsilon(n)) = L_a$ does not imply $\sum_{i=1}^nb_n=\frac{L_a+\mu(n)}{L_1 + \varepsilon(n) }$ $\endgroup$
    – user317176
    Commented Oct 20, 2021 at 5:48

1 Answer 1


$ \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.

  • $\begingroup$ 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. $\endgroup$ Commented Oct 20, 2021 at 6:00
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    $\begingroup$ @AlannRosas Probably, yeah. But I still wanted to point out that skipping over too many steps is not ok when writing proofs. $\endgroup$
    – 5xum
    Commented Oct 20, 2021 at 6:01
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    $\begingroup$ @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 $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i(L_1+\epsilon(n))$ $\endgroup$
    – shintuku
    Commented Oct 20, 2021 at 6:10
  • 1
    $\begingroup$ @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 $n$. $\endgroup$
    – 5xum
    Commented Oct 20, 2021 at 6:11
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    $\begingroup$ @Koro Exactly, $\varepsilon(n)$ is simply some sequence that converges to $0$. In fact, that is all we know of $\varepsilon(n)$. $\endgroup$
    – 5xum
    Commented Oct 20, 2021 at 6:12

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