This problem involves determining the number of terms required for the partial sum of a geometric series to approximate the sum to infinity with a given accuracy.

Problem Setup:

The series is: $2 + \frac{1}{3} + \frac{1}{18} + \frac{1}{108} + \cdots$

The general term of a geometric series is: $a_n = ar^{n-1},$ where $a$ is the first term and $r$ is the common ratio.

Given:

• First term ($a$) = 2,
• Common ratio ($r$) = $\frac{1}{6}$,
• The sum to infinity differs from the partial sum by less than $10^{-5}$.

Step 1: Sum to infinity formula

The sum to infinity for a geometric series is: $S_\infty = \frac{a}{1 - r},$ where $|r| < 1$.

Substituting the values: $S_\infty = \frac{2}{1 - \frac{1}{6}} = \frac{2}{\frac{5}{6}} = \frac{12}{5} = 2.4.$

Step 2: Error tolerance condition

The error ($E$) between the sum to infinity and the partial sum $S_n$ is given by: $E = S_\infty - S_n,$ where $S_n$ is the sum of the first $n$ terms.

For a geometric series, the error can also be expressed as: $E = \frac{a r^n}{1 - r}.$

We are given that: $E < 10^{-5}.$

Substitute $a = 2$, $r = \frac{1}{6}$, and $1 - r = \frac{5}{6}$: $\frac{2 \left(\frac{1}{6}\right)^n}{\frac{5}{6}} < 10^{-5}.$

Simplify: $\frac{2}{5} \cdot \left(\frac{1}{6}\right)^n < 10^{-5}.$

Multiply through by 5: $2 \cdot \left(\frac{1}{6}\right)^n < 5 \cdot 10^{-5}.$

Simplify: $\left(\frac{1}{6}\right)^n < 2.5 \cdot 10^{-5}.$

Step 3: Solve for $n$

Take the natural logarithm ($\ln$) of both sides: $\ln\left(\left(\frac{1}{6}\right)^n\right) < \ln(2.5 \cdot 10^{-5}).$

Using the property of logarithms $\ln(a^b) = b\ln(a)$: $n \ln\left(\frac{1}{6}\right) < \ln(2.5 \cdot 10^{-5}).$

Evaluate $\ln\left(\frac{1}{6}\right)$: $\ln\left(\frac{1}{6}\right) = \ln(1) - \ln(6) = -\ln(6) \approx -1.7918.$

Substitute this into the equation: $n (-1.7918) < \ln(2.5) + \ln(10^{-5}).$

Evaluate $\ln(2.5) \approx 0.9163$ and $\ln(10^{-5}) = -5\ln(10) = -5(2.3026) = -11.513$: $n (-1.7918) < 0.9163 - 11.513.$

Simplify: $n (-1.7918) < -10.5967.$

Divide through by $-1.7918$ (note the sign change when dividing by a negative): $n > \frac{-10.5967}{-1.7918}.$

Simplify: $n > 5.915.$

Since $n$ must be an integer, round up: $n = 6.$

Final Answer:

At least 6 terms are needed for the sum of the series to differ from the sum to infinity by less than $10^{-5}$.

Would you like further details or clarification? 😊

Related Questions:

1. What is the general formula for the partial sum of a geometric series?
2. How does the common ratio $r$ affect the convergence of the series?
3. Can a geometric series converge if $|r| \geq 1$?
4. What is the relationship between the sum to infinity and partial sums?
5. How does increasing the accuracy ($10^{-5}$ to $10^{-6}$) affect the number of terms required?

Tip:

For faster calculations, logarithmic properties like $\ln(ab) = \ln(a) + \ln(b)$ and $\ln(a^b) = b\ln(a)$ are extremely helpful!