If `f(x)= \sum_{n=1}^{\infty }(sin nx)/(4^(n)) and int_(0)^(pi)f(x) dx="log" ((m)/(n))`, then the value of `(m+n)` is ….

Answer

To solve the problem, we need to evaluate the integral of the function \( f(x) = \sum_{n=1}^{\infty} \frac{\sin(nx)}{4^n} \) from \( 0 \) to \( \pi \) and express it in the form \( \log\left(\frac{m}{n}\right) \).

### Step 1: Write the integral of \( f(x) \)

We start with the integral:
\[ \int_0^{\pi} f(x) \, dx = \int_0^{\pi} \sum_{n=1}^{\infty} \frac{\sin(nx)}{4^n} \, dx \]

### Step 2: Interchange the summation and the integral

Assuming we can interchange the summation and the integral (justified by uniform convergence), we have:
\[ \int_0^{\pi} f(x) \, dx = \sum_{n=1}^{\infty} \frac{1}{4^n} \int_0^{\pi} \sin(nx) \, dx \]

### Step 3: Evaluate the integral \( \int_0^{\pi} \sin(nx) \, dx \)

The integral of \( \sin(nx) \) over \( [0, \pi] \) is given by:
\[ \int_0^{\pi} \sin(nx) \, dx = \left[-\frac{1}{n} \cos(nx)\right]_0^{\pi} = -\frac{1}{n} (\cos(n\pi) - \cos(0)) = -\frac{1}{n} (-1 - 1) = \frac{2}{n} \]

### Step 4: Substitute back into the summation

Substituting this result back, we get:
\[ \int_0^{\pi} f(x) \, dx = \sum_{n=1}^{\infty} \frac{1}{4^n} \cdot \frac{2}{n} = 2 \sum_{n=1}^{\infty} \frac{1}{n \cdot 4^n} \]

### Step 5: Evaluate the series \( \sum_{n=1}^{\infty} \frac{1}{n \cdot 4^n} \)

This series can be recognized as the Taylor series expansion for \( -\log(1-x) \) evaluated at \( x = \frac{1}{4} \):
\[ \sum_{n=1}^{\infty} \frac{x^n}{n} = -\log(1-x) \quad \text{for } |x| < 1 \]
Thus,
\[ \sum_{n=1}^{\infty} \frac{1}{n \cdot 4^n} = -\log\left(1 - \frac{1}{4}\right) = -\log\left(\frac{3}{4}\right) = \log\left(\frac{4}{3}\right) \]

### Step 6: Substitute back to find the integral

Now substituting back, we have:
\[ \int_0^{\pi} f(x) \, dx = 2 \cdot \log\left(\frac{4}{3}\right) = \log\left(\left(\frac{4}{3}\right)^2\right) = \log\left(\frac{16}{9}\right) \]

### Step 7: Relate to the given form \( \log\left(\frac{m}{n}\right) \)

From the problem statement, we have:
\[ \log\left(\frac{m}{n}\right) = \log\left(\frac{16}{9}\right) \]
This implies \( m = 16 \) and \( n = 9 \).

### Step 8: Find \( m+n \)

Finally, we calculate:
\[ m+n = 16 + 9 = 25 \]

Thus, the value of \( m+n \) is \( \boxed{25} \).