Evaluate `lim_(nrarrinfty)((1)/(2n+1)+(1)/(2n+2)+......+(1)/(6n))`.

Answer

To evaluate the limit

\[ \lim_{n \to \infty} \left( \frac{1}{2n+1} + \frac{1}{2n+2} + \ldots + \frac{1}{6n} \right), \]

we can rewrite the expression in summation form:

1. **Rewrite the sum**: The sum can be expressed as:

\[ \sum_{r=1}^{4n} \frac{1}{2n + r}. \]

2. **Recognize the limit as a Riemann sum**: As \( n \to \infty \), this summation can be interpreted as a Riemann sum for the integral of a function. Specifically, we can relate it to the integral of \( \frac{1}{2x + 1} \) over the interval from \( 0 \) to \( 4 \):

\[ \lim_{n \to \infty} \sum_{r=1}^{4n} \frac{1}{2n + r} \cdot \frac{1}{n} \approx \int_0^4 \frac{1}{2x + 1} \, dx. \]

3. **Change of variables**: To convert the sum into an integral, we can set \( x = \frac{r}{n} \), which gives us \( r = nx \) and \( dr = n \, dx \). Thus, the sum becomes:

\[ \frac{1}{n} \sum_{r=1}^{4n} \frac{1}{2n + r} \approx \int_0^4 \frac{1}{2n + nx} \, dx = \int_0^4 \frac{1}{n(2 + x)} \, dx. \]

4. **Evaluate the integral**: The integral simplifies to:

\[ \int_0^4 \frac{1}{2 + x} \, dx. \]

5. **Calculate the integral**: The integral can be computed as follows:

\[ \int \frac{1}{2+x} \, dx = \ln |2+x| + C. \]

Evaluating from \( 0 \) to \( 4 \):

\[ \left[ \ln(2+x) \right]_0^4 = \ln(6) - \ln(2) = \ln\left(\frac{6}{2}\right) = \ln(3). \]

6. **Final result**: Therefore, the limit evaluates to:

\[ \lim_{n \to \infty} \left( \frac{1}{2n+1} + \frac{1}{2n+2} + \ldots + \frac{1}{6n} \right) = \ln(3). \]