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 \(R_n<\int_n^\infty\frac{1}{3^x}dx\)</span>

\(\therefore\int_n^\infty\frac{1}{3^x}dx=\lim_{b\rightarrow\infty}\int_n^b\frac{1}{3^x}dx\)

\(=\lim_{b\rightarrow\infty}[-\frac{1}{\ln(3)3^x}]_n^b\)

\(=\lim_{b\rightarrow\infty}[-\frac{1}{\ln(3)3^b}+\frac{1}{\ln(3)3^n}]\)

\(=0+\frac{1}{\ln(3)3^n}\)

\(=\frac{1}{\ln(3)3^n}\)

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 \(R_n<10^{-3}\)</span>

\(\therefore\frac{1}{\ln(3)3^n}<\frac{1}{10^3}\)</span>

\(\ln(3)>\ln(1000)-\ln(\ln(3))\)

\(3^n>\frac{1000}{\ln(3)}\)

\(n\ln(3)>\ln(1000)-\ln(\ln(3))\)

\(n>\frac{3-\ln(\ln(3))}{\ln(3)}\)

\(n>2.645\)

(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

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

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 \(R_n<\int_n^\infty\frac{1}{3^x}dx\)</span>

\(\therefore\int_n^\infty\frac{1}{3^x}dx=\lim_{b\rightarrow\infty}\int_n^b\frac{1}{3^x}dx\)

\(=\lim_{b\rightarrow\infty}[-\frac{1}{\ln(3)3^x}]_n^b\)

\(=\lim_{b\rightarrow\infty}[-\frac{1}{\ln(3)3^b}+\frac{1}{\ln(3)3^n}]\)

\(=0+\frac{1}{\ln(3)3^n}\)

\(=\frac{1}{\ln(3)3^n}\)

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 \(R_n<10^{-3}\)</span>

\(\therefore\frac{1}{\ln(3)3^n}<\frac{1}{10^3}\)</span>

\(\ln(3)>\ln(1000)-\ln(\ln(3))\)

\(3^n>\frac{1000}{\ln(3)}\)

\(n\ln(3)>\ln(1000)-\ln(\ln(3))\)

\(n>\frac{3-\ln(\ln(3))}{\ln(3)}\)

\(n>2.645\)

(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

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