Problem 18
Question
Determine whether the series converges or diverges. $$\sum \frac{1}{k}\left(\frac{1}{\ln k}\right)^{3 / 2}$$
Step-by-Step Solution
Verified Answer
The given series \(\sum \frac{1}{k}\left(\frac{1}{\ln k}\right)^{3 / 2}\) converges by applying the integral test and evaluating the improper integral of its general term, which results in a finite value.
1Step 1: Identify the general term
The general term of the series is:
\(a_k = \frac{1}{k}\left(\frac{1}{\ln k}\right)^{3 / 2}\)
Step 2: Apply the integral test
2Step 2: Apply the integral test
Evaluate the improper integral:
\(\int_{2}^{\infty} \frac{1}{x}\left(\frac{1}{\ln x}\right)^{3 / 2} dx\)
We need to apply a substitution method to solve the integral. Let \(u = \ln x\), then \(du = \frac{1}{x} dx\). Hence, the integral becomes:
\(\int_{\ln2}^{\infty} \frac{1}{u^{3 / 2}} du\)
Step 3: Evaluate the integral
3Step 3: Evaluate the integral
Evaluate the improper integral:
\(\int_{\ln2}^{\infty} \frac{1}{u^{3 / 2}} du = \lim_{m \to \infty} \int_{\ln2}^{m} \frac{1}{u^{3 / 2}} du\)
Since \(\frac{1}{u^{3 / 2}} = u^{-3/2}\), after integration, we get:
\(- 2u^{-1/2}\)
Now, let's apply the limit:
\(\lim_{m \to \infty} (-2m^{-1/2} - (- 2(\ln2)^{-1/2}))\)
As \(m\) goes to infinity, \(m^{-1/2}\) goes to zero:
\(-2(0) + 2(\ln2)^{-1/2} = 2(\ln2)^{-1/2}\)
Step 4: Conclusion
4Step 4: Conclusion
Since the integral is finite (\(2(\ln2)^{-1/2}\)), by the integral test, the series converges:
\(\sum \frac{1}{k}\left(\frac{1}{\ln k}\right)^{3 / 2}\) converges.
Key Concepts
Integral TestSubstitution MethodImproper IntegralsNatural Logarithm Transformations
Integral Test
The integral test is a method used to determine the convergence or divergence of an infinite series. It involves converting the series into an integral and analyzing its behavior. In the context of our exercise, we have an infinite series, \[ \sum \frac{1}{k}\left(\frac{1}{\ln k}\right)^{3 / 2} \]The integral test functions under the following conditions: the series \(f(k)\) must be continuous, positive, and decreasing for all \(k\geq N\) for some positive integer \(N\). The test then evaluates the corresponding integral:\[ \int_{N}^{\infty} f(x) \, dx \]If this integral converges to a finite value, then the original series converges. Conversely, if the integral diverges, the series also diverges. In our example, converting the series into an integral helped us successfully check the series' convergence.
Substitution Method
The substitution method is an invaluable tool in calculus for simplifying integration, particularly when the integral's terms are complex or unwieldy. By substituting part of the integral with a single variable, you can often transform a difficult problem into a more manageable one.
In our exercise, we encounter the integral:\[ \int_{2}^{\infty} \frac{1}{x}\left(\frac{1}{\ln x}\right)^{3 / 2} dx \]This can be intimidating to solve in its original form. However, by letting \(u = \ln x\), it simplifies our work considerably. The differential \(du = \frac{1}{x} dx\) replaces the complicated relationship involving \(x\) and transforms the integral into:\[ \int_{\ln 2}^{\infty} \frac{1}{u^{3 / 2}} du \]The new expression is much simpler to evaluate, demonstrating how powerful substitution can be in tackling integral problems.
In our exercise, we encounter the integral:\[ \int_{2}^{\infty} \frac{1}{x}\left(\frac{1}{\ln x}\right)^{3 / 2} dx \]This can be intimidating to solve in its original form. However, by letting \(u = \ln x\), it simplifies our work considerably. The differential \(du = \frac{1}{x} dx\) replaces the complicated relationship involving \(x\) and transforms the integral into:\[ \int_{\ln 2}^{\infty} \frac{1}{u^{3 / 2}} du \]The new expression is much simpler to evaluate, demonstrating how powerful substitution can be in tackling integral problems.
Improper Integrals
An improper integral is encountered in cases where the integration limits are infinite or when the integrand becomes unbounded within the limits. Handling such integrals requires special techniques, particularly when determining convergence or divergence.
In our example, the improper integral is:\[ \int_{2}^{\infty} \frac{1}{x}\left(\frac{1}{\ln x}\right)^{3 / 2} dx \]After substitution, it transforms into:\[ \int_{\ln2}^{\infty} \frac{1}{u^{3/2}} du \]To solve this, we approach it using limits, evaluating:\[ \lim_{m \to \infty} \int_{\ln2}^{m} \frac{1}{u^{3/2}} du \]This technique helps in finding whether the integral settles to a finite value (indicating convergence) or grows indefinitely (indicating divergence). Our result showed a finite value, confirming the series converges.
In our example, the improper integral is:\[ \int_{2}^{\infty} \frac{1}{x}\left(\frac{1}{\ln x}\right)^{3 / 2} dx \]After substitution, it transforms into:\[ \int_{\ln2}^{\infty} \frac{1}{u^{3/2}} du \]To solve this, we approach it using limits, evaluating:\[ \lim_{m \to \infty} \int_{\ln2}^{m} \frac{1}{u^{3/2}} du \]This technique helps in finding whether the integral settles to a finite value (indicating convergence) or grows indefinitely (indicating divergence). Our result showed a finite value, confirming the series converges.
Natural Logarithm Transformations
Natural logarithm transformations help simplify complex expressions, specifically in calculus problems involving growth rates. The transformation can be leveraged to assess behavior across a range of values.
In our exercise, the series is complicated by the term \(\left(\frac{1}{\ln k}\right)^{3/2}\). By transforming the expression using the natural logarithm, we manage to create a more straightforward problem which is easier to integrate.
This transformation allows the substitution \(u = \ln x\), which effectively removes the complexity and dependencies involving \(x\) from \[ \frac{1}{x}\left(\frac{1}{\ln x}\right)^{3 / 2} \] to \[ \frac{1}{u^{3/2}} \]Using natural logarithms in transformations is especially common when dealing with exponential growth or decay scenarios, making them a staple in calculus revisions and problem-solving but greatly simplifying our series convergence testing.
In our exercise, the series is complicated by the term \(\left(\frac{1}{\ln k}\right)^{3/2}\). By transforming the expression using the natural logarithm, we manage to create a more straightforward problem which is easier to integrate.
This transformation allows the substitution \(u = \ln x\), which effectively removes the complexity and dependencies involving \(x\) from \[ \frac{1}{x}\left(\frac{1}{\ln x}\right)^{3 / 2} \] to \[ \frac{1}{u^{3/2}} \]Using natural logarithms in transformations is especially common when dealing with exponential growth or decay scenarios, making them a staple in calculus revisions and problem-solving but greatly simplifying our series convergence testing.
Other exercises in this chapter
Problem 18
Find a series expansion for the expression. $$\frac{x}{1+4 x^{2}} \text { for }|x|
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Assume that \(f\) is a function with \(\left|f^{(n)}(x)\right| \leq 1\) for all \(n\) and all real \(x\). (The sine and cosine functions have this property.) Es
View solution Problem 18
Test these series for (a) absolute convergence, (b) conditional convergence. \(2^{-}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\).
View solution Problem 18
Expand \(f(x)\) in powers of \(x\) $$f(x)=\cosh x \sinh x$$
View solution