$$\mathop {\lim }\limits_{n \to \infty } {{{a_n}} \over {{b_n}}} = 1$$ 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?

  • 2
    $\begingroup$ $\sum 1/n$ diverges and yet $\lim_{n\to\infty} 1/n=0$, so your proof does not work. $\endgroup$ – Andrés E. Caicedo Dec 5 '13 at 21:37
  • 1
    $\begingroup$ Yes, the statement "If $\sum a_n$ diverges, then $\lim a_n\neq 0$" is false. $\endgroup$ – Thomas Andrews Dec 5 '13 at 21:38
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    $\begingroup$ 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. $\endgroup$ – derivative Dec 5 '13 at 21:46
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    $\begingroup$ @derivative Only if the $a_n$ are eventually positive. $\endgroup$ – Andrés E. Caicedo Dec 5 '13 at 23:12
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    $\begingroup$ @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. $\endgroup$ – Stefan Smith Dec 6 '13 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.

  • $\begingroup$ (Thus, the hypothesis that $a_n,b_n\geqslant 0$ is essential.) $\endgroup$ – Pedro Tamaroff Dec 6 '13 at 16:37
  • $\begingroup$ @PedroTamaroff: this hypothesis would indeed ensure the equivalence, but it was not part of the question. $\endgroup$ – Siméon Dec 6 '13 at 16:48
  • $\begingroup$ Sure. Mine is just a complementary side comment. =) $\endgroup$ – Pedro Tamaroff Dec 6 '13 at 16:50
  • $\begingroup$ How did come up with this example? I'd be glad to know :) $\endgroup$ – Daniel Gagnon Dec 6 '13 at 17:36
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    $\begingroup$ @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. $\endgroup$ – Siméon Dec 6 '13 at 18:25

I suppose you wanted to write that


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 $\sum_{n}^{\infty} a_n$

  • $\begingroup$ 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 $b_n$. $\endgroup$ – Stefan Smith Dec 6 '13 at 2:40
  • $\begingroup$ I just noticed that you yourself gave a comment to the question noting a counterexample for sign-changing series: math.stackexchange.com/questions/30539/… $\endgroup$ – Stefan Smith Dec 6 '13 at 2:41

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