5
$\begingroup$

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

$\endgroup$
  • 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
  • 3
    $\begingroup$ 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
  • 3
    $\begingroup$ @derivative Only if the $a_n$ are eventually positive. $\endgroup$ – Andrés E. Caicedo Dec 5 '13 at 23:12
  • 1
    $\begingroup$ @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
10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ (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
  • 1
    $\begingroup$ @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
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.