Suppose that $(a_n)$ is a sequence of real numbers such that the series
$$\sum_{n=1}^\infty\frac{a_n}n$$is convergent. Show that the sequence
$$b_n=\frac1n\sum^n_{j=1}a_j$$is convergent and find its limit.
1. Given that the series $\sum_{n=1}^\infty \frac{a_n}{n}$ is convergent, we need to show that the sequence $b_n = \frac{1}{n} \sum_{j=1}^n a_j$ is convergent and find its limit.

2. Define $s_n := \sum_{i=1}^n \frac{a_i}{i}$ for $n \geq 0$ and let $s := \lim_{n \to \infty} s_n$. Since $\sum_{n=1}^\infty \frac{a_n}{n}$ is convergent, $s_n$ converges to some limit $s$.

3. We can express $b_n$ as follows:
   \[
   b_n = \frac{1}{n} \sum_{j=1}^n a_j
   \]

4. Notice that:
   \[
   a_j = j(s_j - s_{j-1})
   \]
   Therefore,
   \[
   \sum_{j=1}^n a_j = \sum_{j=1}^n j(s_j - s_{j-1})
   \]

5. Using the above expression, we can rewrite $b_n$:
   \[
   b_n = \frac{1}{n} \sum_{j=1}^n j(s_j - s_{j-1})
   \]

6. We can split the sum:
   \[
   b_n = \frac{1}{n} \left( \sum_{j=1}^n j s_j - \sum_{j=1}^n j s_{j-1} \right)
   \]

7. Notice that the second sum can be rewritten by shifting the index:
   \[
   \sum_{j=1}^n j s_{j-1} = \sum_{i=0}^{n-1} (i+1) s_i = \sum_{i=0}^{n-1} i s_i + \sum_{i=0}^{n-1} s_i
   \]

8. Therefore,
   \[
   b_n = \frac{1}{n} \left( \sum_{j=1}^n j s_j - \left( \sum_{i=0}^{n-1} i s_i + \sum_{i=0}^{n-1} s_i \right) \right)
   \]

9. Simplifying further, we get:
   \[
   b_n = \frac{1}{n} \left( \sum_{j=1}^n j s_j - \sum_{i=0}^{n-1} i s_i - \sum_{i=0}^{n-1} s_i \right)
   \]

10. Notice that:
    \[
    \sum_{j=1}^n j s_j - \sum_{i=0}^{n-1} i s_i = n s_n
    \]
    and
    \[
    \sum_{i=0}^{n-1} s_i = \sum_{i=0}^{n-1} s_i
    \]

11. Therefore,
    \[
    b_n = s_n - \frac{1}{n} \sum_{i=0}^{n-1} s_i
    \]

12. As $n \to \infty$, $s_n \to s$ and $\frac{1}{n} \sum_{i=0}^{n-1} s_i \to s$ because the average of a convergent sequence converges to the same limit.

13. Hence,
    \[
    b_n \to s - s = 0
    \]

14. Therefore, the sequence $b_n$ converges to $0$.

The final answer is $\boxed{0}$