Find the sum to inifinity of the following series:
`1+2/3+3/(3^(2))+4/(3^(3))+`…………….
Find the sum of the following (22-24) series:

Answer

To find the sum to infinity of the series \( S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \), we can follow these steps:

### Step 1: Identify the Series Type
The series can be identified as an AGP (Arithmetic-Geometric Progression) because the numerators \( 1, 2, 3, 4, \ldots \) are in an arithmetic progression (AP) and the denominators \( 3^0, 3^1, 3^2, 3^3, \ldots \) are in a geometric progression (GP).

### Step 2: Write the Series
We can express the series as:
\[ S = \sum_{n=1}^{\infty} \frac{n}{3^{n-1}} \]

### Step 3: Multiply the Series by a Constant
To simplify the series, we multiply \( S \) by \( \frac{1}{3} \):
\[ \frac{S}{3} = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \ldots \]

### Step 4: Subtract the Two Series
Now, we will subtract \( \frac{S}{3} \) from \( S \):
\[ S - \frac{S}{3} = \left(1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots\right) - \left(\frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \ldots\right) \]

### Step 5: Simplify the Result
This subtraction gives:
\[ S - \frac{S}{3} = 1 + \left(\frac{2}{3} - \frac{1}{3}\right) + \left(\frac{3}{3^2} - \frac{2}{3^2}\right) + \left(\frac{4}{3^3} - \frac{3}{3^3}\right) + \ldots \]
\[ = 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \]

### Step 6: Recognize the New Series
The series \( \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \) is a geometric series with first term \( a = \frac{1}{3} \) and common ratio \( r = \frac{1}{3} \).

### Step 7: Sum the Geometric Series
The sum of an infinite geometric series is given by:
\[ \text{Sum} = \frac{a}{1 - r} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]

### Step 8: Combine Results
Now substituting back, we have:
\[ S - \frac{S}{3} = 1 + \frac{1}{2} \]
\[ S - \frac{S}{3} = \frac{3}{2} \]

### Step 9: Solve for \( S \)
Now, we can solve for \( S \):
\[ S \left(1 - \frac{1}{3}\right) = \frac{3}{2} \]
\[ S \left(\frac{2}{3}\right) = \frac{3}{2} \]
\[ S = \frac{3}{2} \cdot \frac{3}{2} = \frac{9}{4} \]

### Final Answer
Thus, the sum to infinity of the series is:
\[ \boxed{\frac{9}{4}} \]