3
$\begingroup$

Here is a series:

$$ - \frac 2 5 + \frac4 6 - \frac 6 7 + \frac8 8 - \frac{10} 9 +\dots $$

The series is convergent (it says so in the back of the book) but the test for divergence fails:

We have, if $\lim_{n\to\infty}{a_n} \neq 0$, then the series is divergent. In this case we have $$ a_n=(-1)^{n}\frac{2n}{n+4}. $$ The $\lim_{n\to\infty}{a_n} \neq 0$, so the series should be divergent. But I know it is convergent.

What am I doing wrong?

$\endgroup$
  • 1
    $\begingroup$ If a series is convergent the general term goes to zero when $n$ goes to infinity, how can you conclude this? $\endgroup$ – DiegoMath Dec 9 '14 at 16:30
  • $\begingroup$ @DiegoMath I have a theorem in my book $\endgroup$ – khajvah Dec 9 '14 at 16:32
  • 1
    $\begingroup$ The theorem should be quoted in full. If it implies that the series in the OP converges, then the theorem is wrong, a serious problem. If the back of the book says the series converges, that is less serious, merely a typo. $\endgroup$ – André Nicolas Dec 9 '14 at 16:39
  • $\begingroup$ @AndréNicolas Most probably the typo is the case. $\endgroup$ – khajvah Dec 9 '14 at 16:46
  • $\begingroup$ The relevant theorem is that if $\lim_{n\to\infty}a_n\ne 0$, and also if the limit does not exist, then the series $\sum a_n$ diverges. $\endgroup$ – André Nicolas Dec 9 '14 at 16:51
9
$\begingroup$

Well, it does not converge: the term converges to $\pm2$, not to $0$, so at least by the Cauchy convergence criterion the sum does not converge.

$\endgroup$
  • $\begingroup$ The book says it converges. It might be a mistake but it is unlucky. $\endgroup$ – khajvah Dec 9 '14 at 16:34
  • 1
    $\begingroup$ @khajvah: what's the book, and which page? Perhaps, we can figure that out. $\endgroup$ – Ilya Dec 9 '14 at 16:35
  • $\begingroup$ The book is James Stewart's Calculus 7th international edition. It is the problem 3 on page 755. $\endgroup$ – khajvah Dec 9 '14 at 16:43
  • $\begingroup$ It's an error in the book (looking at the page now). $\endgroup$ – Nick D. Dec 9 '14 at 16:50
  • $\begingroup$ @NickD.: I now recall the same error in Stewart was once before the subject of a question on MSE. $\endgroup$ – André Nicolas Dec 9 '14 at 16:54
0
$\begingroup$

I think you're taking the wrong sequence. The Leibnitz criteria states that, basically, you transform the alternating sequence into one that does not alternate signs. Then, if the non-alternating sequence converges to zero, the alternating series converges, which it's not your case since the sequence converges to something else or doesn't converge directly.

$\endgroup$
  • 1
    $\begingroup$ Re: "you transform the alternating sequence into one that does not alternate signs. Then, if the non-alternating sequence converges to zero": This part is not necessary. A sequence $a_n$ converges to zero and if only if the sequence $\left|a_n\right|$ converges to zero. (You may be conflating this with a different test, which states that any series $\sum_{n=1}^{\infty}a_n$ must necessarily converge if $\sum_{n=1}^\infty\left|a_n\right|$ does. That test, incidentally, is not specific to strictly alternating series.) $\endgroup$ – ruakh Dec 9 '14 at 21:36

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.