# True/False: $\mathop {\lim }\limits_{n \to \infty } {{{a_n}} \over {{b_n}}} = 1$ implies $\sum {{a_n},\sum {{b_n}} }$ converge or diverge together.

$$\mathop {\lim }\limits_{n \to \infty } {{{a_n}} \over {{b_n}}} = 1$$ Prove the statement implies $\sum {{a_n},\sum {{b_n}} }$ converge or diverge together.
My guess the statement is true.

if $\sum{{a_n}}$ diverges, then $\mathop {\lim }\limits_{n \to \infty } {a_n} \ne 0$

So, \eqalign{ & \mathop {\lim }\limits_{n \to \infty } {a_n} = L \ne 0 \cr & {{\mathop {\lim }\limits_{n \to \infty } {a_n}} \over {\mathop {\lim }\limits_{n \to \infty } {b_n}}} = 1 \Rightarrow {L \over {\mathop {\lim }\limits_{n \to \infty } {b_n}}} = 1 \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {b_n} \ne 0 \cr}

therefore, $\sum {b_n}$ also diverges.

What I was not managed to do is proving that the two series converges together.
Or maybe the statement is not always true?

• $\sum 1/n$ diverges and yet $\lim_{n\to\infty} 1/n=0$, so your proof does not work. Dec 5, 2013 at 21:37
• Yes, the statement "If $\sum a_n$ diverges, then $\lim a_n\neq 0$" is false. Dec 5, 2013 at 21:38
• Lim $a_n/b_n=1$ then $\exists N$ s.t. $a_n/b_n>1/2$ for $n>N$ this implies $b_n<2a_n$ if $\sum a_n$ converges, so does $\sum b_n$ and vice versa. Dec 5, 2013 at 21:46
• @derivative Only if the $a_n$ are eventually positive. Dec 5, 2013 at 23:12
• @Daniel Gagnon : your assertion that "if $\sum a_n$ diverges, then $\lim_{n\to\infty} \neq 0$" is incorrect, and there is a familiar counterexample. Dec 6, 2013 at 2:44

Surprisingly, this statement is false. For a simple counter-example, consider $$a_n = \frac{(-1)^n}{\sqrt{n}},\quad\text{and}\quad b_n = \frac{(-1)^n}{\sqrt{n}} + \frac{1}{n}$$ The condition $a_n \sim b_n$ holds but $\sum a_n$ is convergent whereas $\sum b_n$ is divergent.

• (Thus, the hypothesis that $a_n,b_n\geqslant 0$ is essential.)
– Pedro
Dec 6, 2013 at 16:37
• @PedroTamaroff: this hypothesis would indeed ensure the equivalence, but it was not part of the question. Dec 6, 2013 at 16:48
• Sure. Mine is just a complementary side comment. =)
– Pedro
Dec 6, 2013 at 16:50
• How did come up with this example? I'd be glad to know :) Dec 6, 2013 at 17:36
• @DanielGagnon: the statement is true for series of positive numbers, so you have to look for an example of alternating series that is convergent but not absolutely convergent. Dec 6, 2013 at 18:25

I suppose you wanted to write that

$1)$

if $\overline \lim(\frac {a_n}{b_n})<+\infty$ and $\sum b_n<+\infty$ then $\sum a_n$ converges too.

$2)$$\underline \lim(\frac {a_n}{b_n})>0 and \sum b_n diverges then \sum a_n diverges too. If a_n,b_n\ge 0 and \lim_{n\rightarrow\infty} \frac{a_n}{b_n}=1 then \exists N_1 such that, \frac{a_n}{b_n}>\frac{1}{2} for n\ge N_1 which is equivalent to \quad$$2a_n>b_n$, for $n\ge N_1$

hence, if $\sum_{n}^{\infty} a_n$ converges, then $\sum_{n}^{\infty} 2a_n>\sum_{n}^{\infty} b_n$ also converges.

Similarly $\exists N_2$ such that, $\frac{a_n}{b_n}<\frac{3}{2}$ for $n\ge N_2$

So $a_n<\frac{3b_n}{2}$

If $\sum_{n}^{\infty} b_n$ converges, then also $\sum_{n}^{\infty} a_n$

• you seem to be assuming that all the $b_n$'s are positive. The OP did not say anything about the signs of $a_n$ or $b_n$. Dec 6, 2013 at 2:40
• I just noticed that you yourself gave a comment to the question noting a counterexample for sign-changing series: math.stackexchange.com/questions/30539/… Dec 6, 2013 at 2:41