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

  • $\begingroup$ 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
  • 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 $\lim_{n\to\infty} \lvert a_n\rvert = \infty$.


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