0
$\begingroup$

Edited:

I'm having trouble finding a counterexample for these incorrect statements:

  1. If $\sum a_k$ converges and $\frac{a_k}{b_k} \to 1$ then $\sum b_k$ converges
  2. If $\lim_{k \to \infty} \frac{a_{k+1}}{a_k} \to L$ where $L > 1$ then the series $\sum a_k$ diverges

In the first one, the condition for convergence here needs $a_k$ and $b_k$ to be strictly positives as stated in the limit form comparison test. But I can't find a suitable negative series which following that statement diverges.

In the second one, the problem is that the absolute value is missing in the limit. I thought that I found a counterexample in $(-1)^{n-1}(-2)^n$ but I realised it diverges.

Any help? Thank you

$\endgroup$
  • $\begingroup$ You don't mean a negative series. You mean a series with terms of different signs. :) (Hint: Will you find a counterexample by taking an absolutely convergent series?) $\endgroup$ – Ted Shifrin Jan 20 '18 at 19:46
  • $\begingroup$ Why do you think 2. fails? $\endgroup$ – zhw. Jan 20 '18 at 20:34
1
$\begingroup$
  • For the first:

    You don't want negative sequences, as the theorem would still apply (since $\sum_n a_n$ converges if and only if $\sum_n (-a_n)$ does).

    So you need at least one of the two sequences to be alternating. The simple is then to fix the first, $(a_n)_n$, to be simple: like $a_n \stackrel{\rm def}{=} \frac{(-1)^n}{n}$, one of the simplest alternating convergent series that come to mind.

    We do have (1) $(a_n)_n$ has alternating sign and (2) $\sum_n a_n$ is convergent. So how to choose $(b_n)_n$ now?

    Well, we need $b_n = a_n + c_n$ and $\sum_n b_n$ divergent, so the $c_n$ term must (1) be such that $c_n = o(a_n)$ and (2) $\sum_n c_n$ diverges. Again, let's go for something simple: $c_n \stackrel{\rm def}{=} \frac{1}{n \ln n}$ does the job.

    Summary: choosing (for $n\geq 2$) $$ a_n = \frac{(-1)^n}{n}, \qquad b_n = \frac{(-1)^n}{n} + \frac{1}{n\ln n} $$ works.

  • For the second: the statement is correct. Namely, it's easy to show that then we cannot have$^{(\dagger)}$ $\lim_{n\to\infty} a_n = 0$, which is a necessary condition for convergence.

    $(\dagger)$ specifically, we then have $\lim_{n\to\infty} \lvert a_n\rvert = \infty$.

$\endgroup$

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.