`lim_(n rarr oo) [(1)/(n+1) + (1)/(n+2) + …+ (1)/(6n)]`=

Options

A. log 4
B. log 6
C. log 2
D. none

Answer

To solve the limit

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

we can express the sum in a more manageable form.

### Step 1: Rewrite the sum
The sum can be rewritten as:

\[ \sum_{r=1}^{5n} \frac{1}{n+r}. \]

This is because the terms range from \(n+1\) to \(6n\), which corresponds to \(r\) going from \(1\) to \(5n\).

### Step 2: Factor out \(n\) from the denominator
We can factor \(n\) out of the denominator:

\[ \sum_{r=1}^{5n} \frac{1}{n+r} = \sum_{r=1}^{5n} \frac{1}{n(1 + \frac{r}{n})} = \frac{1}{n} \sum_{r=1}^{5n} \frac{1}{1 + \frac{r}{n}}. \]

### Step 3: Convert the sum to an integral
As \(n\) approaches infinity, the sum can be approximated by an integral. We can express the sum as a Riemann sum:

\[ \frac{1}{n} \sum_{r=1}^{5n} \frac{1}{1 + \frac{r}{n}} \approx \int_{0}^{5} \frac{1}{1+x} \, dx, \]

where we have made the substitution \(x = \frac{r}{n}\).

### Step 4: Determine the limits of integration
The lower limit corresponds to \(r=1\) which gives \(x = \frac{1}{n} \to 0\) as \(n \to \infty\), and the upper limit corresponds to \(r=5n\) which gives \(x = 5\).

### Step 5: Evaluate the integral
Now we evaluate the integral:

\[ \int_{0}^{5} \frac{1}{1+x} \, dx. \]

The integral of \(\frac{1}{1+x}\) is \(\log(1+x)\), so we compute:

\[ \left[ \log(1+x) \right]_{0}^{5} = \log(6) - \log(1) = \log(6). \]

### Step 6: Conclude the limit
Thus, we find that:

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

Therefore, the final answer is:

\[ \boxed{\log(6)}. \]