Key Concepts and Formulas

• Definite Integral as a Limit of a Sum (Riemann Sum): A limit of a sum of the form $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = a}^{b} {f\left( \frac{r}{n} \right)}$ can be converted into a definite integral $\int_{\alpha}^{\beta} {f(x)\,dx}$.
• Conversion Formula: $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = a}^{b} {f\left( \frac{r}{n} \right)} = \int_{\alpha }^{\beta } {f(x)\,dx}$ where $x = \frac{r}{n}$, $dx = \frac{1}{n}$, $\alpha = \mathop {\lim }\limits_{n \to \infty } \frac{a}{n}$, and $\beta = \mathop {\lim }\limits_{n \to \infty } \frac{b}{n}$.
• Integration of Power Functions: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$).

Step-by-Step Solution

Let the given limit be $L$. $L = \mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ........ + {n \over {{{(2n + 1)}^2}}}} \right]$

Step 1: Rewrite the sum in sigma notation and identify the general term. Our goal is to express the given series as a sum that can be converted into a definite integral. We need to find a general term and the range of summation. The terms are: $\frac{1}{n}$, $\frac{n}{(n+1)^2}$, $\frac{n}{(n+2)^2}$, ..., $\frac{n}{(2n+1)^2}$. We can rewrite the first term as $\frac{n}{n^2} = \frac{n}{(n+0)^2}$. The general form of the terms seems to be $\frac{n}{(n+r)^2}$. Let's examine the range of $r$: For the first term, $\frac{n}{(n+0)^2}$, we have $r=0$. For the last term, $\frac{n}{(2n+1)^2}$, we have $n+r = 2n+1$, which implies $r = n+1$. Thus, the sum can be written as: $L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n+1} {\frac{n}{{{{(n + r)}^2}}}}$ However, for the conversion to a Riemann sum, we typically need the upper limit of the summation to be related to $n$ in a way that $\frac{r}{n}$ covers a range from $0$ to some constant. The presence of $+1$ in $2n+1$ is a slight variation. In the limit as $n \to \infty$, the difference between summing up to $n-1$, $n$, or $n+1$ for the index $r$ (when the term involves $n$ in the denominator) does not change the value of the definite integral. For consistent application of the formula, let's consider the sum to have $n$ terms, which means $r$ goes from $0$ to $n-1$. This is a common simplification in such problems. So, we will consider the sum as: $L = \mathop {\lim }\limits_{n \to \infty } \left[ {{n \over {{{(n + 0)}^2}}} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ........ + {n \over {{{(n + (n - 1))}^2}}}} \right]$ This can be written in sigma notation as: $L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {\frac{n}{{{{(n + r)}^2}}}}$

Step 2: Transform the general term into the form $\frac{1}{n} f\left(\frac{r}{n}\right)$. To use the Riemann sum formula, we need to factor out $\frac{1}{n}$ from each term and express the remaining part as a function of $\frac{r}{n}$. Consider the general term: $\frac{n}{{(n+r)}^2}$. Factor out $n^2$ from the denominator: $\frac{n}{{{{(n + r)}^2}}} = \frac{n}{n^2\left(1 + \frac{r}{n}\right)^2}$ Simplify the expression: $= \frac{1}{n} \cdot \frac{1}{\left(1 + \frac{r}{n}\right)^2}$ Now, the limit of the sum becomes: $L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {\frac{1}{n} \cdot \frac{1}{\left(1 + \frac{r}{n}\right)^2}}$

Step 3: Convert the limit of the sum into a definite integral. We have the sum in the form $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 0}^{n - 1} {f\left( \frac{r}{n} \right)}$, where $f\left( \frac{r}{n} \right) = \frac{1}{\left(1 + \frac{r}{n}\right)^2}$. Thus, the function $f(x)$ is obtained by replacing $\frac{r}{n}$ with $x$: $f(x) = \frac{1}{(1+x)^2}$ Now, we determine the limits of integration:

• Lower limit of integration ($\alpha$): The summation starts with $r=0$. $\alpha = \mathop {\lim }\limits_{n \to \infty } \frac{0}{n} = 0$
• Upper limit of integration ($\beta$): The summation ends with $r=n-1$. $\beta = \mathop {\lim }\limits_{n \to \infty } \frac{n-1}{n} = \mathop {\lim }\limits_{n \to \infty } \left(1 - \frac{1}{n}\right) = 1 - 0 = 1$ Therefore, the limit of the sum can be represented as the definite integral: $L = \int_{0}^{1} {\frac{1}{(1+x)^2}\,dx}$

Step 4: Evaluate the definite integral. We need to compute the integral $\int_{0}^{1} {\frac{1}{(1+x)^2}\,dx}$. Let $u = 1+x$. Then $du = dx$. When $x=0$, $u=1+0=1$. When $x=1$, $u=1+1=2$. The integral becomes: $L = \int_{1}^{2} {\frac{1}{u^2}\,du} = \int_{1}^{2} {u^{-2}\,du}$ Using the power rule for integration: $L = \left[ \frac{u^{-2+1}}{-2+1} \right]_{1}^{2} = \left[ \frac{u^{-1}}{-1} \right]_{1}^{2} = \left[ -\frac{1}{u} \right]_{1}^{2}$ Now, apply the limits of integration: $L = \left(-\frac{1}{2}\right) - \left(-\frac{1}{1}\right) = -\frac{1}{2} + 1 = \frac{1}{2}$

Common Mistakes & Tips

• Incorrectly identifying the general term or the range of summation: Carefully examine the terms to find a pattern and the starting and ending values of the index $r$.
• Forgetting to factor out $\frac{1}{n}$: The conversion to a definite integral requires the specific form $\frac{1}{n} f(\frac{r}{n})$.
• Errors in determining the limits of integration: The limits are found by taking the limit of the lower and upper bounds of the summation index divided by $n$.
• Algebraic mistakes during integration or evaluation: Double-check your calculus and arithmetic.

Summary

The problem asks for the limit of a sum as $n$ approaches infinity. This type of problem can be solved by converting the limit of the sum into a definite integral using the Riemann sum definition. We identified the general term of the series, manipulated it into the required form $\frac{1}{n} f(\frac{r}{n})$, and then determined the function $f(x)$ and the limits of integration. Finally, we evaluated the resulting definite integral to find the value of the limit.

The final answer is $\boxed{\frac{1}{2}}$.