The sum of infinite series : `3/4+5/(36)+7/144+9/400+(11)/900+....oo` is

Options

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

Answer

To find the sum of the infinite series \( S = \frac{3}{4} + \frac{5}{36} + \frac{7}{144} + \frac{9}{400} + \frac{11}{900} + \ldots \), we can analyze the series step by step.

### Step 1: Identify the pattern in the series
The series can be rewritten as:
\[ S = \frac{3}{2^2} + \frac{5}{(2 \cdot 3)^2} + \frac{7}{(3 \cdot 4)^2} + \frac{9}{(4 \cdot 5)^2} + \frac{11}{(5 \cdot 6)^2} + \ldots \]

### Step 2: Generalize the terms
From the series, we can observe that the numerator follows the pattern \( 2n + 1 \) for \( n = 1, 2, 3, \ldots \). The denominator can be expressed as \( (n(n+1))^2 \). Thus, we can write the \( n \)-th term of the series as:
\[ T_n = \frac{2n + 1}{(n(n+1))^2} \]

### Step 3: Rewrite the term
We can break down the term \( T_n \):
\[ T_n = \frac{2n + 1}{(n(n+1))^2} = \frac{2n + 1}{n^2(n+1)^2} \]

### Step 4: Separate the fractions
We can separate \( T_n \) into two parts:
\[ T_n = \frac{2n}{n^2(n+1)^2} + \frac{1}{n^2(n+1)^2} \]
This can be simplified further.

### Step 5: Use partial fractions
Using partial fractions, we can express:
\[ \frac{1}{n^2(n+1)^2} = \frac{A}{n} + \frac{B}{n^2} + \frac{C}{n+1} + \frac{D}{(n+1)^2} \]
Solving for constants \( A, B, C, D \) will help us rewrite the series in a telescoping form.

### Step 6: Telescoping series
After finding the constants, we can rewrite the series in a telescoping manner:
\[ S = \sum_{n=1}^{\infty} \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) \]
This will allow us to see that most terms cancel out.

### Step 7: Calculate the limit
The telescoping nature of the series means that as \( n \) approaches infinity, all terms will cancel except for the first term:
\[ S = 1 \]

### Conclusion
Thus, the sum of the infinite series is:
\[ \boxed{1} \]