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Consider the following infinite series. a. Find the first four terms of the sequence of partial sums. b. Use the results of part (a) to find a formula for \(S_{n}\) c. Find the value of the series. $$\sum_{k=1}^{\infty} \frac{2}{3^{k}}$$

Short Answer

Expert verified
Answer: The value of the given infinite series is \(\frac{1}{3}\).

Step by step solution

01

Write down the general formula for the given infinite series

The given infinite series is: $$\sum_{k=1}^{\infty} \frac{2}{3^{k}}$$ This is the general formula we will work with to find the answers to the exercise.
02

Identify the first four terms of the series

To find the first four terms of the series, we can plug in the values \(k=1,2,3,4\) in the general formula: Term 1: \(\frac{2}{3^1} = \frac{2}{3}\) Term 2: \(\frac{2}{3^2} = \frac{2}{9}\) Term 3: \(\frac{2}{3^3} = \frac{2}{27}\) Term 4: \(\frac{2}{3^4} = \frac{2}{81}\)
03

Find the first four partial sums of the series

A partial sum is the sum of the first n terms of the series. We will now calculate the first four partial sums: \(S_1 =\) Term 1 \(= \frac{2}{3}\) \(S_2 =\) Term 1 + Term 2 \(= \frac{2}{3} + \frac{2}{9} = \frac{8}{9}\) \(S_3 =\) Term 1 + Term 2 + Term 3 \(= \frac{2}{3} + \frac{2}{9} + \frac{2}{27} = \frac{26}{27}\) \(S_4 =\) Term 1 + Term 2 + Term 3 + Term 4 \(= \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \frac{2}{81} = \frac{80}{81}\)
04

Derive the formula for nth partial sum \(S_n\) using the results from Step 3

We can notice a pattern in the partial sums: \(S_1 = \frac{2}{3}\), \(S_2 =\frac{8}{9}\), \(S_3= \frac{26}{27}\), \(S_4=\frac{80}{81}\) The numerators are of the form \(2\cdot3^{k-1}\) and the denominators are of the form \(3^{k}-1\). Hence, the formula for the nth partial sum, \(S_n\), is: $$S_n = \frac{2\cdot3^{n-1}}{3^n - 1}$$
05

Find the value of the infinite series

We will now use the formula for \(S_n\) to find the value of the infinite series. The infinite series converges if the limit as n approaches infinity exists: $$\lim_{n\to\infty} S_n = \lim_{n\to\infty} \frac{2\cdot 3^{n-1}}{3^n - 1}$$ Let's analyze the fractions inside the limit separately: $$\frac{2\cdot 3^{n-1}}{3^n - 1} = \frac{1}{3}\cdot \frac{6\cdot 3^{n-2}}{3^2\cdot(3^{n-2} - \frac{1}{3^2})}$$ As \(n\to\infty\), the term \(\frac{1}{3^2}\) becomes insignificant compared to \(3^{n-2}\). Therefore, the limit can be simplified as: $$\lim_{n\to\infty} \frac{1}{3}\cdot \frac{3^{n-2}}{3^{n-2}}$$ The power in the numerator and denominator are now equal, so the limit becomes: $$\lim_{n\to\infty} \frac{1}{3} = \frac{1}{3}$$ Hence, the value of the given infinite series is \(\frac{1}{3}\).

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