Need to evaluate an inf. series

[ToxicBug]

For a series that is Sigma (n!)/(n^n), I did the ratio test and now I need to prove that limit (n^n)/(n+1)^n < 1, can someone give me a hand on how I can do this?

 

[shmoe]

hint:the limit of \left(\frac{n}{n+1}\right)^{-n} should be very familiar...

 

[ToxicBug]

Never seen it before...

 

[CRGreathouse]

Never seen it before...

Caluclate its numeric value for n=1, n=2, n=10, and n=1000.

 

[shmoe]

Never seen it before...

You've never seen:

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

before?Okie, try a comparison of your original sum with \sum_{n=1}^\infty\frac{1}{2n^2}

 

[ToxicBug]

1/e, you could've just said it... as for your second suggestion, how the hell are the two series similar? Are you joking?

 

[shmoe]

1/e, you could've just said it...

I could have also done the whole problem for you, but what's the point of that? I make no apologies for expecting you to meet me well over half way.

as for your second suggestion, how the hell are the two series similar? Are you joking?

No joke. Though I did mean the 2 to be in the numerator to make it more obvious (it doesn't end up mattering though, the terms in your series decay very fast), namely \sum_{n=1}^\infty\frac{2}{n^2}.

Break the terms of your series up into 'n' factors:

\frac{n!}{n^n}=\left(\frac{1}{n}\right)\left(\frac{2}{n}\right)\ldots\left(\frac{n-1}{n}\right)\left(\frac{n}{n}\right)

Now compare with \frac{2}{n^2}.