For every positive integeer $n>1$, let $k(n)$ the largest positive integer $k$ such that there exists a positive integer $m$ such that $n = m^k$.

Find $$lim_{n \rightarrow \infty} \frac{\sum_{j=2}^{j=n+1}{k(j)}}{n}$$
1. Define \( k(n) \) as the largest positive integer \( k \) such that there exists a positive integer \( m \) with \( n = m^k \). This means \( k(n) \) is the largest exponent for which \( n \) can be written as a perfect power.
2. We need to find the limit:
   \[
   \lim_{n \rightarrow \infty} \frac{\sum_{j=2}^{n+1} k(j)}{n}
   \]
3. To understand the behavior of \( \sum_{j=2}^{n+1} k(j) \), note that \( k(j) \) is at most \( \log_2 j \) because \( j = m^k \) implies \( k \leq \log_2 j \).
4. Consider the sum \( \sum_{j=2}^{n+1} k(j) \). Each \( k(j) \) contributes at least 1 to the sum, so:
   \[
   \sum_{j=2}^{n+1} k(j) \geq n
   \]
5. For the upper bound, note that there are at most \( n^{1/2} \) squares less than \( n \), each contributing at most 2 to the sum, at most \( n^{1/3} \) cubes less than \( n \), each contributing at most 3, and so on. Thus:
   \[
   \sum_{j=2}^{n+1} k(j) \leq \sum_{i=1}^{\log_2 n} i n^{1/i}
   \]
6. This sum can be approximated by:
   \[
   \sum_{i=1}^{\log_2 n} i n^{1/i} \leq n + \sum_{i=2}^{\log_2 n} i n^{1/i}
   \]
7. The dominant term in the sum is \( n \), and the remaining terms grow slower than \( n \). Specifically:
   \[
   \sum_{i=2}^{\log_2 n} i n^{1/i} \leq n^{1/2} (\log_2 n)^2
   \]
8. Combining these bounds, we get:
   \[
   n \leq \sum_{j=2}^{n+1} k(j) \leq n + n^{1/2} (\log_2 n)^2
   \]
9. Dividing by \( n \) and taking the limit as \( n \to \infty \):
   \[
   1 \leq \frac{\sum_{j=2}^{n+1} k(j)}{n} \leq 1 + \frac{n^{1/2} (\log_2 n)^2}{n}
   \]
10. The term \( \frac{n^{1/2} (\log_2 n)^2}{n} \) approaches 0 as \( n \to \infty \), so:
    \[
    \lim_{n \rightarrow \infty} \frac{\sum_{j=2}^{n+1} k(j)}{n} = 1
    \]

The final answer is \( \boxed{ 1 } \)