Find the sum of first n terms of the series `3/(1^(2).2^(2))+5/(2^(2).3^(2))+7/(3^(2).4^(2))+.......+`and deduce the sum of infinity .

Answer

To find the sum of the first \( n \) terms of the series

\[ S_n = \frac{3}{1^2 \cdot 2^2} + \frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \ldots \]

we start by identifying the general term of the series. The \( r \)-th term can be expressed as:

\[ T_r = \frac{2r + 1}{r^2 \cdot (r + 1)^2} \]

### Step 1: Rewrite the general term
We can rewrite \( T_r \) using partial fractions. We want to express \( T_r \) as:

\[ T_r = \frac{A}{r} + \frac{B}{r^2} + \frac{C}{r + 1} + \frac{D}{(r + 1)^2} \]

Multiplying through by the common denominator \( r^2 (r + 1)^2 \) and equating coefficients will help us find \( A, B, C, D \).

### Step 2: Find the partial fractions
After setting up the equation and solving for coefficients, we find:

\[ T_r = \frac{1}{r^2} - \frac{1}{(r + 1)^2} \]

### Step 3: Sum the first \( n \) terms
Now, we can sum the first \( n \) terms:

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

This is a telescoping series. When we expand it, we see that most terms cancel out:

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

### Step 4: Final expression for \( S_n \)
Thus, the sum of the first \( n \) terms is:

\[ S_n = 1 - \frac{1}{(n + 1)^2} \]

### Step 5: Find the sum to infinity
To find the sum to infinity, we take the limit as \( n \) approaches infinity:

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

### Conclusion
The sum of the first \( n \) terms of the series is

\[ S_n = 1 - \frac{1}{(n + 1)^2} \]

and the sum to infinity is

\[ S_{\infty} = 1. \]