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

In summary, the conversation is about determining the convergence of a series using the root test and the ratio test. After applying the tests, it is found that the series converges, and as a double check, it is shown to be less than 2/n^2 using the comparison test. One person mentions a typo in the initial steps, and the conversation ends with a discussion about a famous limit.
  • #1
SpicyPepper
20
0

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 [tex]\frac{(n+1)!}{(n+1)^n} * \frac{n^n}{n!}[/tex]

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

I reduce this, and apply the root test:

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

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

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

= 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.
 
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  • #2
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

[tex]\frac{n^n}{(n+1)^n}=\frac{1}{\left( \frac{n+1}{n} \right)^n}=\frac{1}{\left( 1+\frac{1}{n} \right)^n}[/tex].

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.
 
  • #3
Billy Bob said:
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

[tex]\frac{n^n}{(n+1)^n}=\frac{1}{\left( \frac{n+1}{n} \right)^n}=\frac{1}{\left( 1+\frac{1}{n} \right)^n}[/tex].

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. :frown:
 

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

1. What is absolute convergence?

Absolute convergence is a property of an infinite series, which means that the series converges to a finite value regardless of the order in which the terms are added. In other words, the sum of the series always remains the same, regardless of rearrangement of the terms.

2. What is the ratio test for determining absolute convergence?

The ratio test is a mathematical test used to determine the convergence or divergence of an infinite series. It involves taking the limit of the ratio of the absolute values of consecutive terms in the series. If this limit is less than 1, the series is absolutely convergent.

3. How does the ratio test work for n/n^n?

In this case, the ratio test simplifies to taking the limit of n+1/n as n approaches infinity. This limit is equal to 0, which is less than 1. Therefore, the series n/n^n is absolutely convergent.

4. Can the ratio test be used for all series to determine absolute convergence?

No, the ratio test can only be used for series in which the terms are non-negative and approach 0 as n approaches infinity. Additionally, the series must have a factorial or exponential growth in the denominator for the ratio test to be effective.

5. How is the root test different from the ratio test for determining absolute convergence?

The root test also involves taking the limit as n approaches infinity, but it involves taking the nth root of the absolute value of the nth term in the series. If this limit is less than 1, the series is absolutely convergent. The root test is often used when the terms in the series involve exponentials or powers.

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