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.

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.