💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# Test the series for convergence or divergence.$\displaystyle \sum_{n = 2}^{\infty} \frac {1}{(\ln n)^{\ln n}}$

## convergent

Sequences

Series

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

for this problem. We're going toe rewrite Ellen to the Alan. We'Ll rewrite this natural log in as e to the Ellen of Alan event. Then we're taking that to the power of natural out of end. And then we can rearrange this to write it like e the natural log of n and then take this to the power of Ellen Helin of End. Okay. And again, the exponential function of the log function are in verses of each other. So this is just in. So now we're left with in to the power of natural log natural log of end for in large enough, we know that natural log of natural log of n is eventually going to be larger than to. Because natural log of natural law, Govind is goingto blow up as and goes to infinity. So certainly, at some point it will have to be larger than to Okay. And then this sum there's just the sum that we get when we take our original sum, but then replaced that Ellen of Ellen of n with two. And this thing is going tio converge. So we've shown Is that this thing? If we throw out some finite number of terms, Then we'LL be bounded. Since all of these terms are positive and our Siri's is bounded by something, we know that we must have convergence.

Sequences

Series

Lectures

Join Bootcamp