Determining if the series converges

[jdawg]

Homework Statement

_n=1^∞∑\sqrt[n]{3}*\sqrt[n]{n}

Homework Equations

The Attempt at a Solution

I'm thinking that for this series I should use the nth term divergence test.
I know that the limit of each of them is 1, and the whole series converges to 1, but I can't remember why the limits are 1. Could someone please remind me? :)

 

[jbunniii]

Your typesetting is messed up. Is this the series you are evaluating?

$$\sum_{n=1}^{\infty} \sqrt[n]{3} \sqrt[n]{n}$$

If so, you are correct that $\lim_{n\rightarrow \infty} \sqrt[n]{3} = 1$ and $\lim_{n\rightarrow \infty} \sqrt[n]{n} = 1$.

As you said, this implies that $\lim_{n\rightarrow \infty} \sqrt[n]{3} \sqrt[n]{n} = 1$ (the limit of a product is equal to the product of the limits - it's a good idea to check the proof of this if you aren't sure).

What does that imply about the limit of the series?

 

[jdawg]

Sorry,that is what I meant! Since the limit doesn't approach 0 it diverges? But why does the limit equal 1 when you plug in infinity?

 

[jbunniii]

Regarding why $\lim_{n\rightarrow \infty} \sqrt[n]{3} = 1$:

First note that $\sqrt[n]{3} = 3^{1/n}$. Consider the log of this expression: $\log(3^{1/n})$. Can you find the limit the log as $n \rightarrow \infty$? Can you use the result to find the original limit?

 

[jbunniii]

Sorry,that is what I meant! Since the limit doesn't approach 0 it diverges?

Yes, that's right. The terms do not approach zero, so the series diverges.

But why does the limit equal 1 when you plug in infinity?

See my previous post for a suggestion for $\lim_{n \rightarrow 1} \sqrt[n]{3}$. The other limit, $\lim_{n \rightarrow 1} \sqrt[n]{n}$, can be handled similarly but with a slight complication.

 

[jdawg]

I'm sorry, I still don't quite understand what you mean. I get how you're rewriting the square root, but I got a little lost with the logs.

 

[jbunniii]

I claim that $\lim_{n \rightarrow \infty} \log a_n = \log \lim_{n \rightarrow \infty} a_n$ for any convergent sequence of positive values $a_n$. This is because $\log$ is a continuous function. Let's assume we know this result for now - if you're not sure why it's true, we can come back to it later.

So in your case, $a_n = \sqrt[n]{3}$, and we have what I will call Equation 1:
$$\lim_{n \rightarrow \infty} \log \sqrt[n]{3} = \log \lim_{n \rightarrow \infty} \sqrt[n]{3}$$
Working with the left hand side, we have $\log \sqrt[n]{3} = \log (3^{1/n}) = \frac{1}{n} \log 3$ by the basic property of logarithms that $\log x^y = y \log x$.

Now take the limit of that expression: $\lim_{n \rightarrow \infty} \frac{1}{n} \log{3} = ?$ You should be able to get a numerical answer. Once you have that answer, do you see how Equation 1 allows you to find $\lim_{n \rightarrow \infty} \sqrt[n]{3}$?

 

[jdawg]

Okay! Thanks, I think I get it! :)