1. Key Concepts and Formulas

• Definite Integral as a Limit of a Sum (Riemann Sum): The limit of a sum can be represented as a definite integral using the formula: $\lim_{n \to \infty} \sum_{r=a}^{b} f\left(\frac{r}{n}\right) \frac{1}{n} = \int_{a}^{b} f(x) \, dx$ Alternatively, if the summation bounds are expressed in terms of $n$ (e.g., $r$ from $an$ to $bn$): $\lim_{n \to \infty} \sum_{r=an}^{bn} f\left(\frac{r}{n}\right) \frac{1}{n} = \int_{a}^{b} f(x) \, dx$ Here, $x = \frac{r}{n}$, $dx = \frac{1}{n}$, the lower limit of integration is $\lim_{n \to \infty} \frac{\text{lower bound of } r}{n}$, and the upper limit is $\lim_{n \to \infty} \frac{\text{upper bound of } r}{n}$.
• Standard Integral of Inverse Tangent: The integral of $\frac{1}{1+x^2}$ is $\tan^{-1}(x) + C$.

2. Step-by-Step Solution

Step 1: Rewrite the expression and identify the general term. The given expression is: $L = \mathop {\lim }\limits_{n \to \infty } \left( {{n \over {{n^2} + {1^2}}} + {n \over {{n^2} + {2^2}}} + {n \over {{n^2} + {3^2}}} + ..... + {1 \over {5n}}} \right)$ We observe that the first few terms follow the pattern $\frac{n}{n^2+k^2}$ for $k=1, 2, 3, \dots$. Let the general term be $T_r = \frac{n}{n^2+r^2}$.

Step 2: Determine the upper limit of the summation. The last term given is $\frac{1}{5n}$. We need to find the value of $r$ for which $T_r = \frac{1}{5n}$. $\frac{n}{n^2+r^2} = \frac{1}{5n}$ Cross-multiplying gives: $n(5n) = n^2+r^2$ $5n^2 = n^2+r^2$ $4n^2 = r^2$ Taking the positive square root (since $r$ represents the index of summation starting from 1), we get $r = 2n$. Therefore, the sum can be written in sigma notation as: $L = \mathop {\lim }\limits_{n \to \infty } \sum_{r=1}^{2n} \frac{n}{n^2+r^2}$

Step 3: Transform the general term into the Riemann sum form $\frac{1}{n} f\left(\frac{r}{n}\right)$. To convert the sum into an integral, we need to express the general term in the form $\frac{1}{n} f\left(\frac{r}{n}\right)$. We divide the numerator and the denominator of $\frac{n}{n^2+r^2}$ by $n^2$: $\frac{n}{n^2+r^2} = \frac{\frac{n}{n^2}}{\frac{n^2+r^2}{n^2}} = \frac{\frac{1}{n}}{1+\frac{r^2}{n^2}} = \frac{1}{n} \cdot \frac{1}{1+\left(\frac{r}{n}\right)^2}$ Now, the limit becomes: $L = \mathop {\lim }\limits_{n \to \infty } \sum_{r=1}^{2n} \frac{1}{n} \cdot \frac{1}{1+\left(\frac{r}{n}\right)^2}$

Step 4: Determine the limits of integration. We identify $f(x) = \frac{1}{1+x^2}$, where $x = \frac{r}{n}$. The lower limit of integration is $\lim_{n \to \infty} \frac{\text{lower bound of } r}{n} = \lim_{n \to \infty} \frac{1}{n} = 0$. The upper limit of integration is $\lim_{n \to \infty} \frac{\text{upper bound of } r}{n} = \lim_{n \to \infty} \frac{2n}{n} = 2$.

Step 5: Convert the limit of the sum into a definite integral. Using the Riemann sum to definite integral conversion, we have: $L = \int_{0}^{2} \frac{1}{1+x^2} \, dx$

Step 6: Evaluate the definite integral. The integral of $\frac{1}{1+x^2}$ is $\tan^{-1}(x)$. Applying the Fundamental Theorem of Calculus: $L = \left[ \tan^{-1}(x) \right]_{0}^{2}$ $L = \tan^{-1}(2) - \tan^{-1}(0)$ Since $\tan^{-1}(0) = 0$: $L = \tan^{-1}(2) - 0 = \tan^{-1}(2)$

3. Common Mistakes & Tips

• Incorrectly Identifying the Upper Limit: Always verify the upper limit of the summation by equating the general term to the last given term in the series. In this problem, it was crucial to find $r=2n$.
• Missing the $\frac{1}{n}$ Factor: Ensure the expression is manipulated to explicitly show a $\frac{1}{n}$ multiplier outside the function $f(\frac{r}{n})$. This is essential for the conversion to $dx$.
• Confusing Limits of Summation and Integration: Remember that the limits of integration are derived from the limits of the summation index $r$ divided by $n$ as $n \to \infty$.

4. Summary

The problem requires evaluating a limit of a sum, which can be solved by converting it into a definite integral using the concept of Riemann sums. We first identified the general term of the series and determined its upper summation limit by matching it with the last given term. Then, we rewrote the general term in the form $\frac{1}{n} f\left(\frac{r}{n}\right)$. This allowed us to define the function $f(x)$ and determine the limits of integration by considering the behavior of the summation bounds as $n \to \infty$. Finally, we evaluated the resulting definite integral to find the value of the limit.

The final answer is $\tan^{-1}(2)$.

5. Final Answer

The final answer is $\boxed{\text{tan –1 (2)}}$ which corresponds to option (A).