Question

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} 3^{-k}$$

Short answer

Answer: The estimated limit of the sequence of partial sums for the given infinite series is $\frac{1}{2}$.

Step-by-step solution

Write the first four terms of the sequence of partial sums

In order to find the partial sums, we need to sum the first \(n\) terms of the series for \(n=1,2,3,4\). Let's find the first four partial sums \(S_1, S_2, S_3\) and \(S_4\): 1. \(S_1 = 3^{-1} = \frac{1}{3}\) 2. \(S_2 = 3^{-1} + 3^{-2} = \frac{1}{3} + \frac{1}{9} = \frac{4}{9}\) 3. \(S_3 = 3^{-1} + 3^{-2} + 3^{-3} = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} = \frac{13}{27}\) 4. \(S_4 = 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} = \frac{40}{81}\)

Estimate the limit or state that it does not exist

This is a geometric series with a common ratio \(r = \frac{1}{3}\). If \(|r| < 1\), the geometric series converges, and we can use the formula for the sum of an infinite geometric series: $$S_\infty = \frac{a_1}{1-r}$$ Here, the first term \(a_1 = 3^{-1} = \frac{1}{3}\). Since \(|r| = \frac{1}{3} < 1\), the series converges. Now, let's estimate the limit \(S_\infty\): $$S_\infty = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$$ So, the limit of the sequence of partial sums is \(\frac{1}{2}\).