It really doesn't matter because \(\displaystyle \sum_{n=0}^{\infty} \frac{1}{n+1} \) behaves like \(\displaystyle \sum_{n=0}^{\infty} \frac{1}{n} \) by the limit camparison test.

I don't think it can be \(\displaystyle \sum_{n=0}^{\infty} \frac{1}{n} -1 \) because if you make it into an improper fraction, you get \(\displaystyle \sum_{n=0}^{\infty} \frac{1-n}{n} \) which diverges.

Hi p75213...are you a Stargate fan? (if you get that, you are )

Anyway, your problem is in the form of the ratio test for convergence, which basically lets you compare your series to a geometric series to determine it's behavior. I think you'll find your proof here.