sigma n=1 to infinity lnn/n^3 use direct comparison test

To determine the convergence of the series \(\sum_{n=1}^{\infty} \frac{\ln n}{n^3}\) using the **Direct Comparison Test**, we compare it with a simpler series that we know converges or diverges.

### Step 1: Identify a comparable series
We are looking to compare \(\frac{\ln n}{n^3}\) with another series that we know the behavior of. Since \(\ln n\) grows slowly compared to any polynomial, we notice that the term \(\frac{1}{n^2}\) (which is a p-series) might be a good comparison.

### Step 2: Simplify the problem
We know that for large \(n\), \(\ln n\) grows much more slowly than \(n\). Specifically, for \(n\) large enough, \(\ln n \leq n^{0.5}\) (or even a much slower growth), because \(\ln n\) increases very slowly. This gives us an idea that:
\[
\frac{\ln n}{n^3} \leq \frac{n^{0.5}}{n^3} = \frac{1}{n^{2.5}}
\]
for sufficiently large \(n\).

### Step 3: Compare with a p-series
We compare \(\frac{\ln n}{n^3}\) with \(\frac{1}{n^{2.5}}\). The series \(\sum_{n=1}^{\infty} \frac{1}{n^{2.5}}\) is a **p-series** with \(p = 2.5 > 1\), which **converges**.

### Step 4: Apply the Direct Comparison Test
Since \(\frac{\ln n}{n^3} \leq \frac{1}{n^{2.5}}\) for large \(n\), and \(\sum_{n=1}^{\infty} \frac{1}{n^{2.5}}\) converges, by the **Direct Comparison Test**, the given series \(\sum_{n=1}^{\infty} \frac{\ln n}{n^3}\) also **converges**.

### Conclusion:
The series \(\sum_{n=1}^{\infty} \frac{\ln n}{n^3}\) converges by the Direct Comparison Test.

Would you like further details or have any questions?

Here are five related questions to extend your understanding:
1. Can you explain how the growth of \(\ln n\) compares with powers of \(n\) as \(n\) grows large?
2. What is the p-series test, and how is it used to determine convergence?
3. How does the Limit Comparison Test differ from the Direct Comparison Test?
4. What is the difference between absolute convergence and conditional convergence?
5. Can you apply the Direct Comparison Test to \(\sum_{n=1}^{\infty} \frac{1}{n^p}\) for different values of \(p\)?

**Tip:** When using comparison tests, it's important to choose a comparison function that behaves similarly to the given function for large values of \(n\).

Explore the convergence of the series sigma n=1 to infinity ln(n)/n^3 by applying the Direct Comparison Test. Understand how it compares with a p-series and learn step-by-step how the test confirms convergence. Ideal for students studying undergraduate calculus.

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