Question # Consider the following convergent series. a. Find an upper bound for the remainder in terms of n. b. Find how many terms are needed to ensure that the

Series
ANSWERED Consider the following convergent series.
a. Find an upper bound for the remainder in terms of n.
b. Find how many terms are needed to ensure that the remainder is less than $$10^{-3}$$.
c. Find lower and upper bounds (ln and Un, respectively) on the exact value of the series.
$$\sum_{k=1}^\infty\frac{1}{3^k}$$ 2021-02-07

Given,
We are answering the first three subparts as per our honor code.
The series $$\sum_{k=1}^\infty\frac{1}{3^k}$$.Considering the series is convergent then we have to answer the following.
Calculation
(a). Find an upper bound for the remainder in terms of n.
Since we know that $$R_n<\int_n^\infty\frac{1}{3^x}dx$$
$$\therefore\int_n^\infty\frac{1}{3^x}dx=\lim_{b\rightarrow\infty}\int_n^b\frac{1}{3^x}dx$$
$$=\lim_{b\rightarrow\infty}[-\frac{1}{\ln(3)3^x}]_n^b$$
$$=\lim_{b\rightarrow\infty}[-\frac{1}{\ln(3)3^b}+\frac{1}{\ln(3)3^n}]$$
$$=0+\frac{1}{\ln(3)3^n}$$
$$=\frac{1}{\ln(3)3^n}$$
Hence an upper bound for the remainder in terms of n is $$=\frac{1}{\ln(3)3^n}$$
(b) Find how many terms are needed to ensure that the remainder is less than $$10^{-3}$$.
Since given $$R_n<10^{-3}$$
$$\therefore\frac{1}{\ln(3)3^n}<\frac{1}{10^3}$$
$$\ln(3)>\ln(1000)-\ln(\ln(3))$$
$$3^n>\frac{1000}{\ln(3)}$$
$$n\ln(3)>\ln(1000)-\ln(\ln(3))$$
$$n>\frac{3-\ln(\ln(3))}{\ln(3)}$$
$$n>2.645$$
(c). Find lower and upper bounds ($$L_n$$ and $$U_n$$, respectively) on the exact value of the series.
Since $$S_n+\int_{n+1}^\infty\frac{1}{3^x}dx<S<S_n+\int_n^\infty\frac{1}{3^x}dx$$

$$\therefore S_n+\frac{1}{\ln(3)3^{n+1}}<S<S_n+\frac{1}{\ln(3)3^n}$$