Convergence / Divergence of a series

[goraemon]

Homework Statement

Does the following series converge or diverge? $∑\frac{n^5}{n^n}$ (as n begins from 1 and approaches infinity)

Homework Equations

Ratio test?

The Attempt at a Solution

For your reference, thus far I have learned about the geometric series, the limit test, comparison test, a bit about cauchy condensation test, and the ratio test. Thanks for your time.

I tried using the ratio test for this one and ended up getting the following:

$\frac{(n+1)^5}{(n+1)^{n+1}}*\frac{n^n}{n^5}=\frac{n^{n-5}}{(n+1)^{n-4}}=(\frac{n}{n+1})^{n-5}*\frac{1}{n+1}$

(as n approaches infinity)

I'd like to find out if my work is on the right track so far. It certainly seems that the above limit is zero, in which case the original series converges via the ratio test.

Also, please let me know if a different strategy other than the ratio test would've worked out better. Thanks for your time.

 

[Dick]

Homework Statement

Does the following series converge or diverge? $∑\frac{n^5}{n^n}$ (as n begins from 1 and approaches infinity)

Homework Equations

Ratio test?

The Attempt at a Solution

For your reference, thus far I have learned about the geometric series, the limit test, comparison test, a bit about cauchy condensation test, and the ratio test. Thanks for your time.

I tried using the ratio test for this one and ended up getting the following:

$\frac{(n+1)^5}{(n+1)^{n+1}}*\frac{n^n}{n^5}=\frac{n^{n-5}}{(n+1)^{n-4}}=(\frac{n}{n+1})^{n-5}*\frac{1}{n+1}$

(as n approaches infinity)

I'd like to find out if my work is on the right track so far. It certainly seems that the above limit is zero, in which case the original series converges via the ratio test.

Also, please let me know if a different strategy other than the ratio test would've worked out better. Thanks for your time.

I would use the ratio test. And that looks fine. But you should show why that limit is zero. Saying "seems that it does" probably isn't good enough.

 

[benorin]

I would use the comparison test: your general term $\frac{n^5}{n^n}=\frac{1}{n^{n-5}}<\frac{1}{n^2}$, (for sufficiently large n) a general term whose series is convergent.

Maybe I'm giving too much help here, pm me if you think so and I'll erase it...

 

[goraemon]

I would use the ratio test. And that looks fine. But you should show why that limit is zero. Saying "seems that it does" probably isn't good enough.

Ok right...so clearly $\frac{1}{n+1}$ approaches zero as n approaches infinity...which leaves the other part of the equation:

$(\frac{n}{n+1})^{n-5}$

I'm not sure how to find the limit of this equation as n approaches infinity...since the fraction part of it (without the exponent) approaches 1, does it mean the limit is just 1 as well?

In which case the limit of the whole thing would be 1 * 0 = 0?

 

[goraemon]

I would use the comparison test: your general term $\frac{n^5}{n^n}=\frac{1}{n^{n-5}}<\frac{1}{n^2}$, (for sufficiently large n) a general term whose series is convergent.

Maybe I'm giving too much help here, pm me if you think so and I'll erase it...

Thanks for this! Sometimes I fail to see the forest for the trees...

 

[Dick]

Ok right...so clearly $\frac{1}{n+1}$ approaches zero as n approaches infinity...which leaves the other part of the equation:

$(\frac{n}{n+1})^{n-5}$

I'm not sure how to find the limit of this equation as n approaches infinity...since the fraction part of it (without the exponent) approaches 1, does it mean the limit is just 1 as well?

In which case the limit of the whole thing would be 1 * 0 = 0?

No, that limit isn't zero. The limit has the form $1^\infty$. That indeterminant. You can either take the log and use l'Hopital on it, or you could write it in terms of a limit you might recognize (1+1/n)^n. What's the value of that?

 

[Ray Vickson]

Homework Statement

Does the following series converge or diverge? $∑\frac{n^5}{n^n}$ (as n begins from 1 and approaches infinity)

Homework Equations

Ratio test?

The Attempt at a Solution

For your reference, thus far I have learned about the geometric series, the limit test, comparison test, a bit about cauchy condensation test, and the ratio test. Thanks for your time.

I tried using the ratio test for this one and ended up getting the following:

$\frac{(n+1)^5}{(n+1)^{n+1}}*\frac{n^n}{n^5}=\frac{n^{n-5}}{(n+1)^{n-4}}=(\frac{n}{n+1})^{n-5}*\frac{1}{n+1}$

(as n approaches infinity)

I'd like to find out if my work is on the right track so far. It certainly seems that the above limit is zero, in which case the original series converges via the ratio test.

Also, please let me know if a different strategy other than the ratio test would've worked out better. Thanks for your time.

Even easier: let $t_n = n^5/n^n$, so $L_n \equiv \ln(t_n) = 5 \ln(n) - n \ln(n)$. For $n > 7$ we have $L_n < -2 \ln(n)$, hence $t_n < \exp(-2 \ln(n)) = 1/n^2$, so $t_n$ converges, by the comparison test.

 

[goraemon]

No, that limit isn't zero. The limit has the form $1^\infty$. That indeterminant. You can either take the log and use l'Hopital on it, or you could write it in terms of a limit you might recognize (1+1/n)^n. What's the value of that?

OK wait, the value of (1+1/n)^n is the definition of $e$, I think? Which means...
$(\frac{n}{n+1})^{n-5}=\frac{1}{e}$ as n goes to infinity?

So then, continuing from the original post...
$Lim_{n\rightarrow\infty}[(\frac{n}{n+1})^{n-5}*\frac{1}{n+1}]$ becomes $\frac{1}{e}*0=0$?

 

[Dick]

OK wait, the value of (1+1/n)^n is the definition of $e$, I think? Which means...
$(\frac{n}{n+1})^{n-5}=\frac{1}{e}$ as n goes to infinity?

So then, continuing from the original post...
$Lim_{n\rightarrow\infty}[(\frac{n}{n+1})^{n-5}*\frac{1}{n+1}]$ becomes $\frac{1}{e}*0=0$?

Yes, that's it.