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The problem asks to prove that if \( \sum u_n \) is a convergent series of positive real numbers, then \( \sum \frac{u_n}{n} \) is also convergent.

### Proof Outline:

1. **Given:** \( \sum u_n \) is convergent, meaning:
   \[
   \sum_{n=1}^{\infty} u_n = L \quad \text{for some finite} \ L.
   \]
   Since \( u_n > 0 \) for all \( n \), we have \( u_n \to 0 \) as \( n \to \infty \).

2. **Objective:** Prove that \( \sum \frac{u_n}{n} \) is convergent.

3. **Strategy:** We will use the comparison test, comparing \( \frac{u_n}{n} \) with \( u_n \).

### Step-by-Step Proof:

- Let \( u_n \) be the positive terms of the convergent series \( \sum u_n \). Since \( \sum u_n \) converges, the terms \( u_n \to 0 \) as \( n \to \infty \).

- Now consider the series \( \sum \frac{u_n}{n} \). Since \( n \) increases, the terms \( \frac{u_n}{n} \) decay faster than \( u_n \). Specifically, we have:
  \[
  \frac{u_n}{n} \leq u_n \quad \text{for all} \ n \geq 1.
  \]
  Since \( \sum u_n \) converges, by the **comparison test**, if \( \frac{u_n}{n} \) is less than or equal to \( u_n \), and \( \sum u_n \) converges, then \( \sum \frac{u_n}{n} \) also converges.

Thus, by the comparison test, \( \sum \frac{u_n}{n} \) is convergent.

### Conclusion:
If \( \sum u_n \) is a convergent series of positive real numbers, then \( \sum \frac{u_n}{n} \) is also convergent.

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Let me know if you'd like more details or have any questions!

Here are some related questions to consider:
1. What is the **comparison test** for series?
2. How does the **ratio test** work in determining convergence of a series?
3. Can the **root test** be applied to series involving terms like \( \frac{u_n}{n} \)?
4. What happens if \( u_n \) contains both positive and negative terms in this context?
5. Can the convergence of \( \sum \frac{u_n}{n^2} \) be derived from \( \sum u_n \)'s convergence?

**Tip:** In proving convergence of series, using the comparison test is often useful when comparing terms that are "smaller" or "similar" to a known convergent series.

If Σu_n is a convergent series of positive real numbers, prove that Σu_n/n is convergent.

Explore a detailed proof of the convergence of Σu_n/n when Σu_n is a convergent series of positive real numbers. This solution involves using the comparison test to show how the decay of terms leads to convergence. Suitable for students studying advanced calculus or real analysis.

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