Sum up the series
`(2)/(3) + (5)/(9) + (8)/(27)+(11)/(81) + ..... ` to n terms and hence find the sum to infinity .
Sum up to infinity given that x `lt ` numerically .

Answer

To find the sum of the series

\[ S = \frac{2}{3} + \frac{5}{9} + \frac{8}{27} + \frac{11}{81} + \ldots \]

up to \( n \) terms and then find the sum to infinity, we can follow these steps:

### Step 1: Identify the pattern in the series

The series can be rewritten in terms of \( n \):
- The numerators are \( 2, 5, 8, 11, \ldots \) which can be expressed as \( 3n - 1 \) for \( n = 1, 2, 3, \ldots \).
- The denominators are \( 3, 9, 27, 81, \ldots \) which can be expressed as \( 3^n \).

Thus, we can express the \( n \)-th term of the series as:

\[ a_n = \frac{3n - 1}{3^n} \]

### Step 2: Write the sum of the series up to \( n \) terms

The sum of the series up to \( n \) terms is given by:

\[ S_n = \sum_{k=1}^{n} \frac{3k - 1}{3^k} \]

We can split this into two separate sums:

\[ S_n = \sum_{k=1}^{n} \frac{3k}{3^k} - \sum_{k=1}^{n} \frac{1}{3^k} \]

### Step 3: Calculate the first sum \( \sum_{k=1}^{n} \frac{3k}{3^k} \)

This can be simplified as:

\[ \sum_{k=1}^{n} \frac{3k}{3^k} = 3 \sum_{k=1}^{n} \frac{k}{3^k} \]

The sum \( \sum_{k=1}^{n} \frac{k}{r^k} \) can be computed using the formula:

\[ \sum_{k=1}^{n} kx^k = x \frac{d}{dx} \left( \sum_{k=0}^{n} x^k \right) \]

where \( x = \frac{1}{3} \). The sum \( \sum_{k=0}^{n} x^k \) is a geometric series:

\[ \sum_{k=0}^{n} x^k = \frac{1 - x^{n+1}}{1 - x} \]

Differentiating this gives:

\[ \sum_{k=1}^{n} kx^k = x \frac{d}{dx} \left( \frac{1 - x^{n+1}}{1 - x} \right) \]

After differentiating and simplifying, we can find \( \sum_{k=1}^{n} k \left( \frac{1}{3} \right)^k \).

### Step 4: Calculate the second sum \( \sum_{k=1}^{n} \frac{1}{3^k} \)

This is a simple geometric series:

\[ \sum_{k=1}^{n} \frac{1}{3^k} = \frac{\frac{1}{3}(1 - \left(\frac{1}{3}\right)^n)}{1 - \frac{1}{3}} = \frac{1}{2} \left(1 - \left(\frac{1}{3}\right)^n\right) \]

### Step 5: Combine the results to find \( S_n \)

Now combine both sums:

\[ S_n = 3 \sum_{k=1}^{n} \frac{k}{3^k} - \frac{1}{2} \left(1 - \left(\frac{1}{3}\right)^n\right) \]

### Step 6: Find the sum to infinity

To find the sum to infinity, we consider the limit as \( n \to \infty \). The term \( \left(\frac{1}{3}\right)^n \) approaches 0 as \( n \) becomes very large.

Thus, the sum to infinity \( S_{\infty} \) can be calculated as:

\[ S_{\infty} = S_n \text{ as } n \to \infty \]

### Final Result

After evaluating the limits and simplifying, we find:

\[ S_{\infty} = \frac{7}{4} \]