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 following formula: $\lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} f\left(\frac{r}{n}\right) = \int_0^1 f(x) \, dx$ More generally, if the sum is from $r_{min}$ to $r_{max}$, the limits of integration are $a = \lim_{n \to \infty} \frac{r_{min}}{n}$ and $b = \lim_{n \to \infty} \frac{r_{max}}{n}$.

• Integral of $\frac{1}{x}$: The indefinite integral of $\frac{1}{x}$ is $\log_e|x| + C$.

• Fundamental Theorem of Calculus: To evaluate a definite integral $\int_a^b f(x) \, dx$, we find an antiderivative $F(x)$ of $f(x)$ and compute $F(b) - F(a)$.

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Step-by-Step Solution

1. Express the Given Sum in Sigma Notation The given limit is: $L = \mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {1 + n}} + {1 \over {2 + n}} + {1 \over {3 + n}}\, + \,...\, + \,{1 \over {2n}}} \right]$ We need to express this sum in a compact sigma notation. Let's examine the terms: The first term is $\frac{1}{n+1}$. The second term is $\frac{1}{n+2}$. The third term is $\frac{1}{n+3}$. ... The last term is $\frac{1}{2n}$. We can rewrite $\frac{1}{2n}$ as $\frac{1}{n+n}$.

We can see a pattern where the denominator is $n$ plus a variable term that goes from $1$ to $n$. Let $r$ be this variable term. Then the general term is $\frac{1}{n+r}$. When $r=1$, we get $\frac{1}{n+1}$. When $r=2$, we get $\frac{1}{n+2}$. ... When $r=n$, we get $\frac{1}{n+n} = \frac{1}{2n}$. So, the sum can be written as: $\sum_{r=1}^{n} \frac{1}{n+r}$ The limit we need to evaluate is: $L = \mathop {\lim }\limits_{n \to \infty } \sum_{r=1}^{n} \frac{1}{n+r}$ Explanation: Expressing the sum in sigma notation helps us identify the general term and the range of summation, which are crucial for converting it into a definite integral.

2. Transform the Sum into the Standard Riemann Sum Form To use the formula for the definite integral as a limit of a sum, we need to manipulate the general term $\frac{1}{n+r}$ into the form $\frac{1}{n} f\left(\frac{r}{n}\right)$. We can rewrite the general term as: $\frac{1}{n+r} = \frac{1}{n\left(1 + \frac{r}{n}\right)}$ Now, we can separate the $\frac{1}{n}$ factor: $\frac{1}{n+r} = \frac{1}{n} \cdot \frac{1}{1 + \frac{r}{n}}$ Substituting this back into the limit expression: $L = \mathop {\lim }\limits_{n \to \infty } \sum_{r=1}^{n} \left[ \frac{1}{n} \cdot \frac{1}{1 + \frac{r}{n}} \right]$ Explanation: By factoring out $\frac{1}{n}$ and isolating the $\frac{r}{n}$ term, we prepare the sum to match the standard Riemann sum structure, where $\frac{r}{n}$ will become our integration variable $x$.

3. Convert the Limit of the Sum to a Definite Integral We now compare our transformed sum with the Riemann sum formula: $\lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} f\left(\frac{r}{n}\right) = \int_a^b f(x) \, dx$ From our expression, $\frac{1}{n} \cdot \frac{1}{1 + \frac{r}{n}}$, we can identify:

• The function $f(x)$ corresponds to $\frac{1}{1+x}$, since $\frac{r}{n}$ is replaced by $x$.
• The $\frac{1}{n}$ factor outside the function corresponds to $dx$.

Next, we determine the limits of integration, $a$ and $b$:

• Lower Limit ($a$): This is obtained by taking the limit of $\frac{r_{min}}{n}$ as $n \to \infty$. In our sum, $r_{min} = 1$. $a = \lim_{n \to \infty} \frac{1}{n} = 0$
• Upper Limit ($b$): This is obtained by taking the limit of $\frac{r_{max}}{n}$ as $n \to \infty$. In our sum, $r_{max} = n$. $b = \lim_{n \to \infty} \frac{n}{n} = 1$ Therefore, the limit of the sum can be represented by the definite integral: $L = \int_0^1 \frac{1}{1+x} \, dx$ Explanation: This step is the core of the problem. We have successfully transformed the given limit of an infinite sum into a definite integral by identifying the integrand and the integration interval.

4. Evaluate the Definite Integral Now we evaluate the definite integral: $L = \int_0^1 \frac{1}{1+x} \, dx$ The antiderivative of $\frac{1}{1+x}$ is $\log_e|1+x|$. Using the Fundamental Theorem of Calculus: $L = \left[ \log_e |1+x| \right]_0^1$ Substitute the upper and lower limits: $L = \log_e |1+1| - \log_e |1+0|$ $L = \log_e 2 - \log_e 1$ Explanation: We apply the standard integration formula for $\frac{1}{u}$ and then use the Fundamental Theorem of Calculus to evaluate the integral at the limits.

5. Simplify the Result Since $\log_e 1 = 0$: $L = \log_e 2 - 0$ $L = \log_e 2$ Explanation: The final simplification of the evaluated integral gives us the value of the limit.

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Summary

The problem asks for the limit of a sum, which can be effectively solved by converting it into a definite integral using the concept of Riemann sums. We first express the given sum in sigma notation, then manipulate the general term to fit the form $\frac{1}{n} f\left(\frac{r}{n}\right)$. This allows us to identify the function $f(x)$ and the limits of integration. Evaluating the resulting definite integral $\int_0^1 \frac{1}{1+x} \, dx$ yields $\log_e 2$.

The final answer is $\boxed{{\log _e}2}$.