Convergence and divergence help

[kidsmoker]

Homework Statement

Determine whether the following converges or diverges:

\sum\frac{n^{n}}{(n+1)^{(n+1)}}

with the sum going from n=1 to n=infinity.

Homework Equations

Comparison Test, Ratio Test.

The Attempt at a Solution

This should be do-able with the above two tests, since those are the only ones covered in the book up to this point. I think it diverges but I can't seem to get anywhere with either test. Any help appreciated!

 

[randomlogic78]

I think you are going to want to use the ratio test... try setting it up this way...

\lim_{n\rightarrow\infty} \frac{\frac{(n+1)^{n+1}}{((n+1)+1)^{((n+1)+1)}}}{\frac{n^n}{(n+1)^{(n+1)}}}

 

[Dick]

You can try that. But it's going to give you a limit of 1. So the test is inconclusive. What you really need to do is find a way to use the famous limit (1+1/n)^n=e.

 

[gabbagabbahey]

The ratio test is inconclusive (check for yourself using randomlogic's setup); why don't you show us what you've tried for the comparison test?

 

[randomlogic78]

Indeed it is inconclusive.

 

[Dick]

Indeed it is inconclusive.

Reasonable suggestion, nonetheless.

 

[kidsmoker]

Okay thanks, I think i got it:


\frac{n^{n}}{(n+1)^{n+1}} = \frac{1}{(n+1)}(\frac{n}{n+1})^{n} = \frac{1}{(n+1)}\frac{1}{(\frac{n+1}{n})^{n}} = \frac{1}{(n+1)}\frac{1}{(1+\frac{1}{n})^{n}}} \geq \frac{1}{(n+1)}\frac{1}{(e+1)}

using the fact that (1+1/n)^n tends to e.

Now \frac{1}{e+1}\sum{\frac{1}{n+1}} diverges, therefore our series must diverge too :-)


Does that seem okay?

 

[Dick]

Okay thanks, I think i got it:


\frac{n^{n}}{(n+1)^{n+1}} = \frac{1}{(n+1)}(\frac{n}{n+1})^{n} = \frac{1}{(n+1)}\frac{1}{(\frac{n+1}{n})^{n}} = \frac{1}{(n+1)}\frac{1}{(1+\frac{1}{n})^{n}}} \geq \frac{1}{(n+1)}\frac{1}{(e+1)}

using the fact that (1+1/n)^n tends to e.

Now \frac{1}{e+1}\sum{\frac{1}{n+1}} diverges, therefore our series must diverge too :-)


Does that seem okay?

That's it. (1+1/n)^n is actually <=e, isn't it? But there's nothing wrong with playing it safe and saying <=(e+1) either.