\( \lim _ { m \rightarrow + \infty } \frac { \Lambda } { \sqrt { 1 + m ^ { 2 } } } + \frac { \Lambda } { \sqrt { 2 + m ^ { 2 } } } + \frac { \Lambda } { \sqrt { 3 + n ^ { 2 } } } + \cdots + \frac { \Lambda } { \sqrt { n + m ^ { 2 } } } \)

To evaluate the limit: \[ \lim_{m \rightarrow +\infty} \left( \frac{\Lambda}{\sqrt{1 + m^2}} + \frac{\Lambda}{\sqrt{2 + m^2}} + \frac{\Lambda}{\sqrt{3 + m^2}} + \cdots + \frac{\Lambda}{\sqrt{n + m^2}} \right) \] we can analyze each term in the sum individually as \( m \) approaches infinity. ### Step-by-Step Analysis: 1. **Understanding Each Term:** Each term in the sum is of the form: \[ \frac{\Lambda}{\sqrt{k + m^2}} \] where \( k \) is a constant ranging from 1 to \( n \). 2. **Simplifying for Large \( m \):** For large \( m \), the term \( k \) becomes negligible compared to \( m^2 \), so: \[ \sqrt{k + m^2} \approx \sqrt{m^2} = m \] Therefore, each term simplifies to: \[ \frac{\Lambda}{\sqrt{k + m^2}} \approx \frac{\Lambda}{m} \] 3. **Summing Up All Terms:** Since there are \( n \) such terms in the sum, the entire expression approximates to: \[ \sum_{k=1}^{n} \frac{\Lambda}{m} = \frac{n\Lambda}{m} \] 4. **Taking the Limit as \( m \rightarrow +\infty \):** As \( m \) grows without bound, the expression \( \frac{n\Lambda}{m} \) approaches 0: \[ \lim_{m \rightarrow +\infty} \frac{n\Lambda}{m} = 0 \] ### Conclusion: Therefore, the limit of the given expression as \( m \) approaches infinity is: \[ 0 \] **Answer:** 0