# 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?

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$\sum 1/n$ diverges and yet $\lim_{n\to\infty} 1/n=0$, so your proof does not work. –  Andres Caicedo Dec 5 '13 at 21:37
Yes, the statement "If $\sum a_n$ diverges, then $\lim a_n\neq 0$" is false. –  Thomas Andrews Dec 5 '13 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. –  derivative Dec 5 '13 at 21:46
@derivative Only if the $a_n$ are eventually positive. –  Andres Caicedo Dec 5 '13 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. –  Stefan Smith Dec 6 '13 at 2:44

## 3 Answers

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.

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(Thus, the hypothesis that $a_n,b_n\geqslant 0$ is essential.) –  Pedro Tamaroff Dec 6 '13 at 16:37
@PedroTamaroff: this hypothesis would indeed ensure the equivalence, but it was not part of the question. –  Siméon Dec 6 '13 at 16:48
Sure. Mine is just a complementary side comment. =) –  Pedro Tamaroff Dec 6 '13 at 16:50
How did come up with this example? I'd be glad to know :) –  Daniel Gagnon Dec 6 '13 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. –  Siméon Dec 6 '13 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 $\sum_{n}^{\infty} a_n$

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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 $b_n$. –  Stefan Smith Dec 6 '13 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/… –  Stefan Smith Dec 6 '13 at 2:41