Summations converging and diverging.

niyati · #1 $Old$ November 14th, 2007, 08:16 AM

I need to find out if:

Summation of n = 1 to infinity of n^n/n!

converges or diverges. If it converges, what is the sum?

I tried the ratio test, since it has a factorial in it, and I end up getting ((n+1)/n)^n, where I apply L'Hopital's Rule, by taking the natural log, pulling the n forward and whatnot, but I'm not sure where to go from there. I mean, I'm sure that I can make it a fraction, but taking limit as n goes to infinity of ln((n+1)/n)/(1/n), and since it's in the 0/0 form, I can take the derivatives, and then take e to that answer, and if it's less than one, it converges. Greater than one is divergence, and equal to one is inconclusive.

However, would the ratio found be the sum that the problem is looking for (if it converges)?

Am I approaching this correctly?

Jhevon · #2 $Old$ November 14th, 2007, 09:04 AM

Quote:

Originally Posted by niyati $View Post$

I need to find out if:

Summation of n = 1 to infinity of n^n/n!

converges or diverges. If it converges, what is the sum?

I tried the ratio test, since it has a factorial in it, and I end up getting ((n+1)/n)^n, where I apply L'Hopital's Rule, by taking the natural log, pulling the n forward and whatnot, but I'm not sure where to go from there. I mean, I'm sure that I can make it a fraction, but taking limit as n goes to infinity of ln((n+1)/n)/(1/n), and since it's in the 0/0 form, I can take the derivatives, and then take e to that answer, and if it's less than one, it converges. Greater than one is divergence, and equal to one is inconclusive.

However, would the ratio found be the sum that the problem is looking for (if it converges)?

Am I approaching this correctly?

the sum diverges, you can see this pretty quickly since $\lim_{n \to \infty}\frac {n^n}{n!} \ne 0$

but if you want to do it the hard way and do the ratio test, then here is how to proceed.

you ended up with:

$\lim_{n \to \infty} \left| \left( \frac {n + 1}n \right)^n \right| = \lim_{n \to \infty} \left| \left( 1 + \frac 1n \right)^n \right| = |e| = e > 1$

thus the series diverges

in general the ration found is not the sum, as can readily be seen here. the ratio found is e while the actual sum diverges to infinity

niyati · #3 $Old$ November 14th, 2007, 10:23 AM

The divergence you calculated easily was because of the nth term test, right?

Jhevon · #4 $Old$ November 14th, 2007, 10:29 AM

Quote:

Originally Posted by niyati $View Post$

The divergence you calculated easily was because of the nth term test, right?

No. well, i don't know if that's what you call it, but i used the Test for Divergence.

recall the theorem:

Theorem: If $\sum a_n$ converges, then $\lim a_n = 0$

the contrapositive of this theorem is:

The Test For Divergence: If $\lim a_n \ne 0$ , then $\sum a_n$ diverges

niyati · #5 $Old$ November 14th, 2007, 10:36 AM

yep! That's it.

Thank you!