Problem 8
Question
Consider the infinite series \(\sum_{k=1}^{\infty} \frac{1}{k} .\) Evaluate the first four terms of the sequence of partial sums.
Step-by-Step Solution
Verified Answer
Answer: The first four terms of the sequence of partial sums are \(1, 1.5, 1.833,\) and \(2.083\).
1Step 1: Understanding the sequence of partial sums
A sequence of partial sums is a sequence where each term is the sum of the first n terms of a series. For our case, we need to find the first four partial sums for the series \(\sum_{k=1}^{\infty} \frac{1}{k}\). So, we will find the sum of the first, second, third, and fourth terms of the series respectively.
2Step 2: Finding the first partial sum (S1)
The first partial sum is simply the first term of the series when k=1. So, plug in k=1 into the series formula:
\(S_1 = \frac{1}{1} = 1\)
3Step 3: Finding the second partial sum (S2)
For the second partial sum, we will sum the first two terms of the series:
\(S_2 = \frac{1}{1} + \frac{1}{2} = 1 + 0.5 = 1.5\)
4Step 4: Finding the third partial sum (S3)
Now, for the third partial sum, we sum the first three terms of the series:
\(S_3 = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = 1 + 0.5 + \frac{1}{3} \approx 1.833\)
5Step 5: Finding the fourth partial sum (S4)
Finally, for the fourth partial sum, we sum the first four terms of the series:
\(S_4 = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = 1 + 0.5 + \frac{1}{3} + 0.25 \approx 2.083\)
So, the first four terms of the sequence of partial sums for the given infinite series are \(1, 1.5, 1.833,\) and \(2.083\).
Other exercises in this chapter
Problem 8
Evaluate each geometric sum. $$\sum_{k=0}^{10}\left(\frac{1}{4}\right)^{k}$$
View solution Problem 8
Explain how two sequences that differ only in their first ten terms can have the same limit.
View solution Problem 9
Use the Divergence Test to determine whether the following series diverge or state that the test is inconclusive. $$\sum_{k=0}^{\infty} \frac{k}{2 k+1}$$
View solution Problem 9
Evaluate each geometric sum. $$\sum_{k=0}^{20}\left(\frac{2}{5}\right)^{2 k}$$
View solution