Absolute convergence: ratio/root test n/n^n

[SpicyPepper]

Homework Statement

Doing some problems from textbook, I need to determine whether the series is absolutely convergent, conditionally convergent, or divergent.

n!/n^n

I plugged it into WA, and it says the series doesn't converge, but I'm not sure how to figure it out.

Homework Equations

The Attempt at a Solution

First, I applied the root test

lim n->inf \frac{(n+1)!}{(n+1)^n} * \frac{n^n}{n!}

lim n->inf \frac{(n+1)n!}{(n+1)(n+1)^n} * \frac{n^n}{n!}

I reduce this, and apply the root test:

lim n->inf \sqrt[n]{\frac{n^n}{(n+1)^n}}

lim n->inf \frac{n}{n+1}

lim n->inf \frac{1}{1 + 1/n}

= 1

1 means that it's inconclusive. I'm not sure if I applied the tests incorrectly or if I'm supposed to try something else.

 

[Billy Bob]

It seems you took the root of the ratio. That's wrong. Don't combine the two tests. Use one or the other.

The ratio test will work. (Your first step has a typo, but the second step has fixed it.) To finish it off, observe

\frac{n^n}{(n+1)^n}=\frac{1}{\left( \frac{n+1}{n} \right)^n}=\frac{1}{\left( 1+\frac{1}{n} \right)^n}.

The last expression has a famous limit.

Actually the series converges, and as a double check using the comparison test, it is less than 2/n^2.

 

[SpicyPepper]

It seems you took the root of the ratio. That's wrong. Don't combine the two tests. Use one or the other.

The ratio test will work. (Your first step has a typo, but the second step has fixed it.) To finish it off, observe

\frac{n^n}{(n+1)^n}=\frac{1}{\left( \frac{n+1}{n} \right)^n}=\frac{1}{\left( 1+\frac{1}{n} \right)^n}.

The last expression has a famous limit.

Actually the series converges, and as a double check using the comparison test, it is less than 2/n^2.

Thanks for mentioning the typos, I see them. I meant to say I applied the ratio test first, and the exponent in the denominator of my first line should be n+1.

I remember the limit from deriving it with L'Hopital's rule, 1/e. Thanks, I simply didn't see I could reduce it by dividing by n^n/n^n.