Determine the convergence of the telescoping series ∑n=1∞ln(nn+1)\sum_{n=1}^{\infty} \ln\left(\frac{n}{n+1}\right)∑n=1∞ln(n+1n). What is the sum SN=∑n=1Nln(nn+1)S_N = \sum_{n=1}^{N} \ln\left(\frac{n}{n+1}\right)SN=∑n=1Nln(n+1n)?
SN=−ln(N+1)S_N = -\ln(N+1)SN=−ln(N+1); series diverges to −∞-\infty−∞
SN=ln(1N+1)=−ln(N+1)S_N = \ln\left(\frac{1}{N+1}\right) = -\ln(N+1)SN=ln(N+11)=−ln(N+1); series converges to 000
SN=−ln(N+1)S_N = -\ln(N+1)SN=−ln(N+1); series converges to −ln(∞)-\ln(\infty)−ln(∞)
SN=ln(N)S_N = \ln(N)SN=ln(N); series diverges to ∞\infty∞