Answer
(a) If a series is convergent by the integral test , we estimate the sum by integrating the function from $0$ to infinity . For example: $\lim\limits_{n \to \infty}\frac{1}{(n-3)^2}$ can be written as $\int_0^\infty\frac{1}{(n-3)^2}$.
(b) If a series is convergent by the comparison test , we estimate the sum by estimating its compared function. For example: $\sum_{n=1}^\infty\frac{5}{2n^2+4n
+3}$ , at infinity , the series behave like$\sum_{n=1}^\infty\frac{5}{2n^2}$ which is similar to $\sum_{n=1}^\infty\frac{1}{n^2}$. Fro this we know that the sum of the series equal to $\frac{5}{2} \int_1^\infty \frac{1}{x^2}$.
(c) If a series is converegent by Alternating Test , we need to keep adding the terms in that given series untill we reasch the desired accuracy. This is known as Alternating Series Estimation.
Work Step by Step
(a) If a series is convergent by the integral test , we estimate the sum by integrating the function from $0$ to infinity . For example: $\lim\limits_{n \to \infty}\frac{1}{(n-3)^2}$ can be written as $\int_0^\infty\frac{1}{(n-3)^2}$.
(b) If a series is convergent by the comparison test , we estimate the sum by estimating its compared function. For example: $\sum_{n=1}^\infty\frac{5}{2n^2+4n
+3}$ , at infinity , the series behave like$\sum_{n=1}^\infty\frac{5}{2n^2}$ which is similar to $\sum_{n=1}^\infty\frac{1}{n^2}$. Fro this we know that the sum of the series equal to $\frac{5}{2} \int_1^\infty \frac{1}{x^2}$.
(c) If a series is converegent by Alternating Test , we need to keep adding the terms in that given series untill we reasch the desired accuracy. This is known as Alternating Series Estimation.
