Thread: series (converging)

1. series (converging)

Im not sure how to do the part in red, do i use the comparison test?

2. $\displaystyle \displaystyle\sum_{r=1}^{\infty}\log\dfrac{r+1}{r} =\displaystyle\sum_{r=1}^{\infty}(\;\log({r+1})-\log {r}\$

(telecopic series)

Fernando Revilla

Edited: Sorry, I didn't read the first part. So, an alternative is:

$\displaystyle \log \dfrac{r+1}{r}\sim \dfrac{1}{r}\;(r\rightarrow +\infty)$

and the harmonic series is divergent, so the given series is divergent.