I have read different articles on Convergence vs Divergence. How the general formula for the partial sums of the series here and here be calculated?
$$s_n=\sum_{i=2}^n \frac1{i^2-1}=\frac34-\frac1{2n}-\frac1{2(n+1)}$$ $$s_n=\sum_{i=1}^n\frac1{3^{i-1}}=\frac32\left(1-\frac1{3^n}\right)$$
Please let me know how? Thanks sabbir
For the second, this series is geometric, and the formula for finding a partial sum of a geometric series is a(1-r^n)/(1-r) where r is the term being raised to a power of i, and n is the number of terms you evaluate to, and a is the coefficient of the sum, which in this case is 1. So, in the second case, r is 1/3, a is 1, and n is unspecified. With some algebra you can turn (1-1/3^n)/(1-1/3) into the solution they give you. One important step is to make sure i=0, which you can do by changing the exponent from i-1 to just i.
