$$\sum_{n=1}^{\infty} (\ln(2(n+1))- \ln(2n))$$ I was able to plug this into a calculator to determine that the series is divergent. I also graphed the series to observe a decreasing, continuous positive function. Thus, using the integral test seemed like a reasonable choice to determine convergence.
If I didn't have access to a graphing calculator or a calculator at all to solve this problem, how would I go about it? Also, if there are any tools (videos, sites, personal tips and tricks) that can help me determine what test to use to determine convergence/divergence as well as computing the sum, it would be very helpful.
$$\sum_{n=1}^{\infty} (\ln(2(n+1))- \ln(2n)) = \sum_{n=1}^{\infty} \ln\left(\frac{n+1}{n}\right) = \ln\left(\prod_{n=1}^{\infty} \frac{n+1}{n}\right) = \lim_{m \to \infty} (\ln(m+1) - \ln(1))$$
This may also be seen by adding the terms after regrouping as seen by: $$\sum_{n=1}^{\infty} ( \ln(2n+2) - \ln(2n) ) = \lim_{m \to \infty} (\ln(2(m+1)) - \ln(2)) = \lim_{m \to \infty} \ln(m+1).$$
From this the series diverges. One may also use the "comparison test" to show that it diverges.