0
$\begingroup$

As I understand, the two conditions for an alternating series to converge using the Leibniz criterion are:

  1. Absolute value of sequence terms is monotome decreasing and

  2. Limit of the terms of the sequence to infinity is 0.

I initially thought that the second condition must necessarily imply the first condition, so I was wondering why both conditions were stated; however I now appreciate that these two conditions are seperate, although I cannot think of an example that will help me to wrap my head around this.

Does anyone have any examples of an alternating series for which the limit of the sequence formed by the terms is zero but the terms are not monotome decreasing and therefore the series fails the Leibniz criterion and diverges?

$\endgroup$
4
  • $\begingroup$ decreases doesn't mean converges to zero. $\endgroup$ Nov 12, 2016 at 19:11
  • 1
    $\begingroup$ $1-0+1/2-0+1/3-0+ \cdots$ $\endgroup$
    – zhw.
    Nov 12, 2016 at 19:13
  • $\begingroup$ @AbdallahHammam Thank you for your reply. I know condition 1 does not imply condition 2. A simple example would be the sum of k from 1 to infinity of (1+(1/k)), which is indeed decreasing but not tending toward zero. My query was with condition 2 implying condition 1, which I know is not the case but I couldn't think of any specific examples where this was not the case. $\endgroup$
    – Meep
    Nov 12, 2016 at 20:10
  • $\begingroup$ @zhw. Thank you for your reply. This seems like a piece-wise series? Surely the Leibniz criterion doesn't apply to this? $\endgroup$
    – Meep
    Nov 12, 2016 at 20:12

1 Answer 1

2
$\begingroup$

$$ \sum_{n=2}^\infty\frac{(-1)^n}{n^p+(-1)^n} $$ diverges if $0<p\le1/2$ and converges if $p>1/2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .