The value of `\lim_{n \to \infty } sum_(k=1)^n (n-k)/(n^2) cos((4k)/n)` equals to

Options

A. (a) `1/4 sin 4+1/15 cos 4-1/16`
B. (b) `1/4 sin 4-1/16cos 4 +1/16`
C. (c) `1/16(1-sin4)`
D. (d) `1/16(1-cos4)`

Answer

To solve the limit problem, we need to evaluate the expression:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{n-k}{n^2} \cos\left(\frac{4k}{n}\right) \]

### Step 1: Rewrite the summation

First, we can rewrite the term \(\frac{n-k}{n^2}\) as follows:

\[ \frac{n-k}{n^2} = \frac{n}{n^2} - \frac{k}{n^2} = \frac{1}{n} - \frac{k}{n^2} \]

Thus, we can express the limit as:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \left(\frac{1}{n} - \frac{k}{n^2}\right) \cos\left(\frac{4k}{n}\right) \]

### Step 2: Split the summation

Now, we can split the summation into two parts:

\[ \lim_{n \to \infty} \left( \sum_{k=1}^{n} \frac{1}{n} \cos\left(\frac{4k}{n}\right) - \sum_{k=1}^{n} \frac{k}{n^2} \cos\left(\frac{4k}{n}\right) \right) \]

### Step 3: Recognize the Riemann sum

The first summation \(\sum_{k=1}^{n} \frac{1}{n} \cos\left(\frac{4k}{n}\right)\) resembles a Riemann sum for the integral of \(\cos(4x)\) from \(0\) to \(1\) as \(n \to \infty\):

\[ \int_{0}^{1} \cos(4x) \, dx \]

### Step 4: Calculate the first integral

Calculating the integral:

\[ \int \cos(4x) \, dx = \frac{1}{4} \sin(4x) + C \]

Thus, evaluating from \(0\) to \(1\):

\[ \left[\frac{1}{4} \sin(4x)\right]_{0}^{1} = \frac{1}{4} \sin(4) - \frac{1}{4} \sin(0) = \frac{1}{4} \sin(4) \]

### Step 5: Analyze the second summation

Now, consider the second summation:

\[ \sum_{k=1}^{n} \frac{k}{n^2} \cos\left(\frac{4k}{n}\right) \]

This can be rewritten as:

\[ \frac{1}{n^2} \sum_{k=1}^{n} k \cos\left(\frac{4k}{n}\right) \]

As \(n \to \infty\), this also resembles a Riemann sum. We can express it as:

\[ \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k}{n}\right) \cos\left(4 \cdot \frac{k}{n}\right) \to \int_{0}^{1} x \cos(4x) \, dx \]

### Step 6: Calculate the second integral

Using integration by parts where \(u = x\) and \(dv = \cos(4x)dx\):

\[ du = dx, \quad v = \frac{1}{4} \sin(4x) \]

Thus,

\[ \int x \cos(4x) \, dx = x \cdot \frac{1}{4} \sin(4x) - \int \frac{1}{4} \sin(4x) \, dx \]

The integral of \(\sin(4x)\) is:

\[ -\frac{1}{16} \cos(4x) \]

So, we have:

\[ \int x \cos(4x) \, dx = \frac{1}{4} x \sin(4x) + \frac{1}{16} \cos(4x) \]

Evaluating from \(0\) to \(1\):

\[ \left[\frac{1}{4} x \sin(4x) + \frac{1}{16} \cos(4x)\right]_{0}^{1} = \left(\frac{1}{4} \sin(4) + \frac{1}{16} \cos(4)\right) - \left(0 + \frac{1}{16}\right) \]

This simplifies to:

\[ \frac{1}{4} \sin(4) + \frac{1}{16} \cos(4) - \frac{1}{16} \]

### Step 7: Combine results

Now, combining both parts:

\[ \frac{1}{4} \sin(4) - \left(\frac{1}{4} \sin(4) + \frac{1}{16} \cos(4) - \frac{1}{16}\right) \]

This simplifies to:

\[ \frac{1}{16} (1 - \cos(4)) \]

### Final Answer

Thus, the limit evaluates to:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{n-k}{n^2} \cos\left(\frac{4k}{n}\right) = \frac{1}{16} (1 - \cos(4)) \]

### Conclusion

The correct answer is:

\[ \frac{1}{16} (1 - \cos(4)) \]