# Root test fails, now what?

freshman2013

## 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.

Staff Emeritus
Did you try the comparison test?

Mentor

## 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.

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.

Homework Helper
Dearly Missed

## 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.

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

Homework Helper
Dearly Missed
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.

Last edited:
Staff Emeritus