Practice problem
Evaluate the sum ∑n=12019ln(n+1n).\sum^{2019}_{n = 1}\ln\left(\frac{n + 1}{n}\right).n=1∑2019ln(nn+1).
202020202020
12020\frac{1}{2020}20201
−ln(2020)-\ln(2020)−ln(2020)
ln(2020)\ln(2020)ln(2020)
Type a concept, like “slope” or “photosynthesis”.
