# Root test fails, now what?

• freshman2013
That's where I got my idea. I think your idea is worth exploring.What do you mean by "looks somewhat like this"? I don't see any obvious way to apply the comparison test here, and the limit comparison test doesn't seem to work either.In summary, the series (1-1/n^(1/3))^n may or may not converge based on various tests, but the comparison test and limit comparison test do not seem to be useful in this case.

## Homework Statement

determine whether the series (1-1/n^(1/3))^n converge or diverge

## Homework Equations

all the testing procedure

## The Attempt at a Solution

So I did the root test first, but the limit on the inside is 1. I then tried the ratio test but then when I tried taking the limit, I ended up with e^(-infinity+infinity). Finally, I tried the test for divergence on the original series, but i turned out be zero so it might or might not converge. Now I'm totally stuck.

Did you try the comparison test?

freshman2013 said:

## Homework Statement

determine whether the series (1-1/n^(1/3))^n converge or diverge

## Homework Equations

all the testing procedure

## The Attempt at a Solution

So I did the root test first, but the limit on the inside is 1. I then tried the ratio test but then when I tried taking the limit, I ended up with e^(-infinity+infinity).
You should never end up with something like this ( -∞ + ∞), as it is indeterminate. I agree with what you got on the root test, so that's not a useful test, and the n-th term test seems to give a value of 0, so it doesn't tell us anything.

I would take another look at the ratio test to see what the limit actually is.

freshman2013 said:
Finally, I tried the test for divergence on the original series, but i turned out be zero so it might or might not converge. Now I'm totally stuck.

freshman2013 said:

## Homework Statement

determine whether the series (1-1/n^(1/3))^n converge or diverge

## Homework Equations

all the testing procedure

## The Attempt at a Solution

So I did the root test first, but the limit on the inside is 1. I then tried the ratio test but then when I tried taking the limit, I ended up with e^(-infinity+infinity). Finally, I tried the test for divergence on the original series, but i turned out be zero so it might or might not converge. Now I'm totally stuck.

If
$$t_n = \left(1 - \frac{1}{n^{1/3}}\right)^n,$$
you can look at ##L_n = \ln(t_n)## and use the series expansion of ##\ln (1-x)## for small ##x = 1/n^{1/3}##, to get the behavior of ##t_n## for large n. In fact, you can even get a simple upper bound ##u_n##, so that ## 0 < t_n < u_n##, and ##\sum u_n## is easy to analyze.

The OP might not have been exposed to series yet...

Mark44 said:
The OP might not have been exposed to series yet...

Series are not really needed here; it is enough to know that 1-x < exp(-x) for x > 0, so that ln(1-x) < -x for small x > 0.

Anyway, if the OP has seen series the method could be useful; if he/she has not yet seen series it is perhaps not useful. However, I have no way of knowing what the OP has, or has not, seen.

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I'm not seeing how that would help. If it's working toward the ratio test, the ratio of consecutive terms approaches 1 as n→∞, so that test is inconclusive.

There is a very simple series that does converge that looks somewhat like this.