Series Convergence: Is the Root Test Always Reliable?

[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.

 

[D H]

Did you try the comparison test?

 

[Mark44]

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.

 

[Ray Vickson]

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.

 

[Mark44]

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

 

[Ray Vickson]

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.

 

[D H]

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.