Question

Let S be the sum of a series of \(\sum {{a_n}} \) that has shown to be convergent by the Integral Test and let f(x) be the function in that test. The remainder after n terms is

\({R_n} = S - {S_n} = {a_{n + 1}} + {a_{n + 2}} + {a_{n + 3}} + .......\)

Thus, R_n is the error made when S_n , the sum of the first n terms is used as an approximation to the total sum S.

(a) By comparing areas in a diagram like figures 3 and 4 (but with x ≥ n), show that

\(\int\limits_{n + 1}^\infty {f(x)dx \le {R_n} \le \int\limits_n^\infty {f(x)dx} } \)

(b) Deduce from part (a) that

\({S_n} + \int\limits_{n + 1}^\infty {f(x)dx \le S \le {S_n} + \int\limits_n^\infty {f(x)dx} } \)

Short answer

(a) On Solving the equation we get ,\(\int\limits_{n + 1}^\infty {f(x)dx \le {R_n} \le \int\limits_n^\infty {f(x)dx} } \)

(b) On Solving the equation we get ,\({S_n} + \int\limits_{n + 1}^\infty {f(x)dx \le S \le {S_n} + \int\limits_n^\infty {f(x)dx} } \)

Step-by-step solution

(a)Integral Test

Consider \(S = \sum {{a_n}} \) is convergent by the Integral Test, S_n will also be convergent and also,

\(\begin{aligned}{}{R_n} &= S - {S_n}\\{\rm{ }} &= {a_{n + 1}} + {a_{n + 2}} + {a_{n + 3}} + {a_{n + 4}} + .....\end{aligned}\)

Integrating

Since S_nis convergent then

\({S_{n - 1}} \le {S_n} \le {S_{n + 1}}\)

Multiply by -1 both sides

\( - {S_{n - 1}} \le - {S_n} \le - {S_{n + 1}}\)

Add S on both sides

\(S - {S_{n - 1}} \le S + {S_n} \le S - {S_{n + 1}}\)

Integrating both sides we get

\(\)\(\int\limits_{n + 1}^\infty {f(x)dx} \le {R_n} \le \int\limits_n^\infty {f(x)dx} \)

Adding Sn from part (a)

From part (a) we have

\(\int\limits_n^\infty {f(x)dx} \le {R_n} \le \int\limits_n^\infty {f(x)dx} \)

\( \Rightarrow \int\limits_{n + 1}^\infty {f(x)dx} \le S - {S_n} \le \int\limits_n^\infty {f(x)dx{\rm{ Since }}{R_n} = S - {S_n}} \)

Add _n on both sides we get

\({S_n} + \int\limits_{n + 1}^\infty {f(x)dx \le S \le {S_n} + \int\limits_n^\infty {f(x)dx} } \)