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# 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.

### 1 Answer

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- 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.

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