0
$\begingroup$

Ok, so here's the issue. I've got these two alternating series and the problem states that one is conditional and one is absolutely convergent. The problem I'm having is that, using both the root test and the ratio test, I am coming up with both as absolutely convergent. So I must be doing something wrong. Summary of my work below:

enter image description here

Alternating Series (1/2)^n Converges by AST: decreasing, positive, limit of zero.
Ratio test = 1/2 < 1 so absolute convergence.
Root test = 1/2 < 1 so absolute convergence

enter image description here

Alternating series 1/sqrt(n)
Converges by AST: decreasing, positive, limit of zero.
Ratio test = 1 so no conclusion can be drawn Root test = lim of the n root of 1/sqrt(n) <1 so absolutely convergent

There's a mistake somewhere but I'm clearly missing it. Thanks for any help!

$\endgroup$
  • $\begingroup$ Recheck your application of the Root test for the second series. $\endgroup$ – David Mitra Jul 12 '14 at 14:57
1
$\begingroup$

The mistake is that in fact $$\lim_{n\to\infty} (\sqrt{n})^{1/n} = 1,$$ so the root test is also inconclusive. A simple comparison with for example the harmonic series shows that the second is not absolutely convergent.

$\endgroup$
1
$\begingroup$

For the second series and taking the absolute value we find the Riemann series

$$\sum_n\frac1{\sqrt n}$$ which's divergent since $\frac12\le1$. Notice that we can also see the divergence by comparison with the harmonic series $$\frac1{\sqrt n}\ge \frac1n$$

$\endgroup$
0
$\begingroup$

Though they weren't included in my course, there are several more series tests that are mightily helpful when you can't resolve it through the ratio or root test.

There is the divergence test: check whether the limit of the function becomes zero or not, if that limit isn't zero -> you are certain it diverges. If it is zero inconclusive.

Is it a geometric series ($\sum a.r^n$)? Your first example is a geometric series. Then if $-1<r<1$ => converges to $\frac{a}{1-r}$. If $r$ doesn't fall in that range it diverges. Since $-1<1/2<1$ is true you know is absolutely convergent, it converges to -2.

Is it a p series $\sum \frac{1}{n^p}$? If p>1 it converges, if p=<1 it diverges. The second series can be seen as an alternating p series. If we were to just look at the p-series part then we know that if the given series had just been $\frac{1}{n^{1/2}}$ that $1/2 < 1$ and that a simplified form of what you were given would diverge. The alternating part wouldn't make it convergent.

When you recognize a type of series $b_n$ in the given series you can use the direct comparison test or limit comparison test to prove that the given series $a_n$ acts similar to the simplified form.

The direct limit test says that when $a_n < b_n$ for all n then it converges too. If $a_n > b_n$ it diverges too.

The limit comparison test takes the limit of $\frac{a_n}{b_n}$. If that gives you an integer > 0 then you know they act the same.

And lastly there is another useful test: the alternating series test, that you can use for a series containing $(-1)^n$. You check whether the function that follows after $(-1)^n$ can answer the following questions positively. Does $b_n$ decrease in value? And is the limit of $b_n$ 0. If both answers are positive then it converges, otherwise it diverges.

I found that knowing about these other tests greatly increased my insight into the series and understanding on whether they converge and diverge in a way that the root and ratio test don't really. These tests are my last resort tests when the series is much too complicated, and I can only solve it mechanically. A tip on these two: if you see a fractional (the k-permutation sign) then the ratio test is probably your best bet, because the ratio test helps you to get rid of them.

$\endgroup$

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.