Let `a_(n)=int_(0)^(pi//2)(1-sint )^(n) sin 2t,`
then `lim_(n to oo)sum_(n=1)^(n)(a_(n))/(n)`is equal to

Options

A. `1//2`
B. `1`
C. `4//3`
D. `3//2`

Answer

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

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{a_k}{k} \]

where

\[ a_n = \int_{0}^{\frac{\pi}{2}} (1 - \sin t)^n \sin 2t \, dt. \]

### Step 1: Evaluate \( a_n \)

We start by rewriting \( a_n \):

\[ a_n = \int_{0}^{\frac{\pi}{2}} (1 - \sin t)^n \sin 2t \, dt. \]

Using the substitution \( u = 1 - \sin t \), we have:

\[ \sin t = 1 - u \quad \text{and} \quad dt = -\frac{du}{\cos t} = -\frac{du}{\sqrt{1 - (1-u)^2}} = -\frac{du}{\sqrt{u(2-u)}}. \]

The limits change as follows:
- When \( t = 0 \), \( u = 1 \).
- When \( t = \frac{\pi}{2} \), \( u = 0 \).

Thus, we can rewrite the integral:

\[ a_n = \int_{1}^{0} u^n \cdot 2(1-u) \left(-\frac{du}{\sqrt{u(2-u)}}\right). \]

Reversing the limits gives:

\[ a_n = 2 \int_{0}^{1} u^n (1-u) \, du. \]

### Step 2: Calculate the integral

The integral \( \int_{0}^{1} u^n (1-u) \, du \) can be computed using the Beta function:

\[ \int_{0}^{1} u^n (1-u) \, du = B(n+1, 2) = \frac{(n)! (1)!}{(n+1)!} = \frac{1}{n+1}. \]

Thus,

\[ a_n = 2 \cdot \frac{1}{n+1} = \frac{2}{n+1}. \]

### Step 3: Substitute \( a_n \) back into the limit

Now we substitute \( a_n \) back into the limit:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{a_k}{k} = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{2}{(k)(k+1)}. \]

### Step 4: Simplify the summation

We can simplify \( \frac{2}{k(k+1)} \) using partial fractions:

\[ \frac{2}{k(k+1)} = 2 \left(\frac{1}{k} - \frac{1}{k+1}\right). \]

Thus, the sum becomes:

\[ \sum_{k=1}^{n} \left(2 \left(\frac{1}{k} - \frac{1}{k+1}\right)\right) = 2 \left(1 - \frac{1}{n+1}\right) = 2 - \frac{2}{n+1}. \]

### Step 5: Take the limit

Now, we take the limit as \( n \to \infty \):

\[ \lim_{n \to \infty} \left(2 - \frac{2}{n+1}\right) = 2 - 0 = 2. \]

### Final Answer

Thus, the final answer is:

\[ \boxed{2}. \]