0
$\begingroup$

My professor worked this following problem in class, and I'm having some difficulty following the logic.

We had the alternating series, $\sum\limits_{n=2}^{\infty} \frac{(-1)^n \sqrt{n}}{\ln n}$. This fails the alternating series test, as $\lim\limits_{n \to \infty} \frac{\sqrt{n}}{\ln n} = \infty$. He used this as a basis to say that, by the Divergence Test, the series diverges.

I can't follow this, though. The Divergence Test, if I'm not mistaken, is on the entirety of the general term of the series, $\frac{(-1)^n \sqrt{n}}{\ln n}$. The test that he ran should show that the series is not absolutely convergent, though it could still be conditionally convergent. The other problem is that $\lim\limits_{n \to \infty} \frac{(-1)^n \sqrt{n}}{\ln n}$ is not zero because it doesn't exist; the sign alters back and forth. But wouldn't this logic apply to any alternating series, for example, $\sum\limits_{n=1}^{\infty} (-1)^n \sin^{-1} \frac{1}{n}$?

Revision to the above: After running this new series through wolframalpha, it does indeed some to have a limit of zero, though I can't seem to make out why, as $(-1)^n$ doesn't converge, and if we break up the limit of a product into the product of the limits, we get one term that doesn't converge (oscillates between $1$ and $-1$) and one term that goes to $0$. Does the $0$ "win out"?

More compactly, my questions are:

(a) Is it sufficient to run the alternating series test on the absolute value of the general term of the series? Or should we run it on the entire alternating series, and conclude that the oscillation of the $(-1)^n$ factor will prevent the limit from converging?

(b) Is it true that the lim as $n \to \infty$ of the entirety of the general term will always be zero for a series that converges by the alternating-series test? Is this true because the $0$ factor "wins out" against the oscillating factor?

Thanks in advance.

$\endgroup$
  • $\begingroup$ That limit property only holds if both limits exist. $\endgroup$ – Andrew Li May 26 '18 at 0:49
  • $\begingroup$ By definition, we run the alternating series test on an alternating series to show conditional convergence. The test checks the limit of the absolute value of the $n$th term because if the absolute value has a limit of $0$ and decreases, then the alternating series converges (conditionally). $\endgroup$ – Andrew Li May 26 '18 at 0:53
  • $\begingroup$ I think I understand this, though wouldn't we say that the test is inconclusive if $\lim\limits_{n \to \infty} \lvert u_n \rvert \neq 0$ or doesn't exist? My professor seemed to conclude divergence by the Divergence Test, but it seems to me that we only know that the alternating series test would be inconclusive, and would need to then run the Divergence Test on $u_n$, which includes the $(-1)^n$ factor. $\endgroup$ – Matt.P May 26 '18 at 0:59
  • $\begingroup$ If the limit of $|u_n|$ isn't zero, it fails the $n$th term test (or what you may call the divergence test). Then it's not inconclusive, it conclusively tells you that it's divergent because a series with summands that increase indefinitely or do not converge onto $0$ mean in essence that the series' terms never stop increasing... $\endgroup$ – Andrew Li May 26 '18 at 1:01
  • $\begingroup$ Excellent. I didn't know this. Thank you very much. $\endgroup$ – Matt.P May 26 '18 at 1:05
1
$\begingroup$

If $\lim_{n\to+\infty}u_n\ne 0 $ or does not exist then both series $$\sum u_n $$ and $$\sum (-1)^nu_n $$ diverge since the general term does not go to zero.

If $\lim_{n\to+\infty}u_n=0$, and $(u_n) $ is decreasing, the series $$\sum (-1)^nu_n $$ is convergent.

$\endgroup$
  • $\begingroup$ I think I'm a bit confused with notation here, particularly in terms of whether $u_n$ represents the entire general term, with the alternating factor, $(-1)^n$, or whether it represents the "positive" component. If I'm reading what you've written correctly -- please correct me if I'm wrong -- it seems to suggest that if the alternating series test fails and that limit is not 0 or doesn't exist, both the absolute value of the series and the 'alternating' series converge. But, aren't we testing for conditional convergence with the alternating-series test? $\endgroup$ – Matt.P May 26 '18 at 0:57
  • $\begingroup$ If $\lim |u_n|$ is not zero then $\lim u_n $ is not zero. $\endgroup$ – hamam_Abdallah May 26 '18 at 0:59
  • $\begingroup$ @Matt.P If the alternating series test fails, a series is neither conditionally nor absolutely convergent (due to the $n$th term test). $\endgroup$ – Andrew Li May 26 '18 at 0:59
  • $\begingroup$ I didn't know either of these above facts, though they seem to make a lot of sense. It seems I've been incorrectly treating the alternating-series test like, say, the ratio test, which is inconclusive with a ratio of 1. I didn't know it could be a test for divergence as well. Thank you both! $\endgroup$ – Matt.P May 26 '18 at 1:01
1
$\begingroup$

Divergence Test (That You Probably Now): If $\lim_{n} a_n$ does not converge to zero, then $\sum_{n} a_n$ diverges.

However, the divergence test still works if you apply it to absolute value of the terms. That's because $\lim_{n} a_n = 0$ if and only if $\lim_{n} |a_n| = 0$.

Therefore:

Divergence Test (That's Easier to Use): If $\lim_{n} |a_n|$ does not converge to zero, then $\sum_{n} a_n$ diverges.

$\endgroup$
  • $\begingroup$ This is perhaps the most intuitive way to think of this. The 'if and only if' condition you wrote is especially useful. Thank you very much. $\endgroup$ – Matt.P May 26 '18 at 1:28

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.