# Does this sequence converge or diverge?

1. an= [(2^n)(n!) + 1]/(n^n)

At first glance, I thought it would be a convergent sequence because of the (n^n) in the denominator. But after trying and not being able to show its convergence, I compared it to (2/n)^n with the squeeze theorem. (Basically inf > an > (2/n)^n)

I set the limit as n approaches inf, took the natural log and put 1/2n as the denominator. I used the L'Hopital method and got to lim e^((-2/n^2)(n/2)/(-1/n^2)), which canceled out to lim e^n.

I concluded that the sequence diverges. Did I do this correctly? This problem's been bugging me for two days now. Guidance please!

Dick
Homework Helper
It is true your original sequence diverges. But showing it's greater than 2^n/n^n won't prove that, because 2^n/n^n converges to 0. If you take the log(2^n/n^n) you get log(2^n)-log(n^n)=n*log(2)-n*log(n). That's not a ratio you can apply l'Hopital to. So I really don't know how you got that limit. It really goes to -infinity. What I would do is go back to your original sequence and do a ratio test. Find lim a_(n+1)/a_n. If that ratio is greater than 1, it diverges.

Oh, thanks!!

I did (2/n)^n and took the ln of all that. I fail to see how that's not indeterminate (0 to the infinity?)...

This was actually on my exam Monday. I used both the squeeze theorem and ratio test, but I didn't think the ratio test would get me points because I thought that was reserved for series only?

Not sure if I did all the algebra right, but doing the ratio test for (2^(n)n!/n^n) got me 2/e which is less than 1. I remember on the exam I got a value greater than 1.... This discrepancy is strange.

I think focusing on this one problem for so long has warped my super algebra skills. Other finals to study for.

But yeah, I started off on the wrong foot with this problem because of the n^n in the denominator. Tricky I guess :3 Fingers crossed for partial credit.

Dick
Homework Helper
You usually apply the ratio test to series, that's true. But it can also be applied to sequences. Which is pretty obvious if you think about it. Whether you got something larger than 1 or less than 1 depends on whether you were computing an+1/an or an/an+1. I get 2/e for the ratio of an+1/an. OOPS. Wait. I just realize I got it wrong for just that reason. an does converge, sorry about that. lim an+1/an=2/e. But it's still true that showing an is greater than a convergent sequence doesn't show it converges.

...ok. Yes I know it makes sense to do all that sorry I'll go and be sad about that problem now.

Dick
Homework Helper
On the thing about 2^n/n^n. Taking log gives you n*log(2/n). That's not indeterminant. It's infinity*(-infinity). The limit is -infinity. The log of 2^n/n^n goes to -infinity. 2^n/n^n goes to zero.

Dick
Homework Helper
...ok. Yes I know it makes sense to do all that sorry I'll go and be sad about that problem now.

I think you'll do better next time. Don't be that sad.

That was for the final, though. Ahh.. I was told by upperclassmen that I'd feel dread and feel like giving up, and that time has come. College is hard. And misery-inducing. But enough about that problem, I guess I'll just have to see how I did on the rest of the test, which had only 8 problems and took me 3 hours to do.

Thanks for your help, though. I could actually feel my heart fall after realizing I had it compleeetely wrong, that was pretty extreme haha. This will only make winter break sweeter.

Dick