Can a P-series be more complex than Sigma U sub k = k^(-p)?
I found something with a quantity to the (-1/3) power. It threw me off.
Sigma U sub k = (2k-1)^(-1/3)
I said it diverged but i want to see if i worked it out the right way (Integral Test).
Feel free to work it out. Please and thank-you.
- davidLv 69 years agoFavorite Answer
You're right. You can show that
Sigma U sub k from 2 to inf < Integral (1 to inf) of x^(-1/3) dx which is infinite.