Find the sum `Sigma_(n=1)^(oo)(3n^2+1)/((n^2-1)^3)`

Explanation

To find the sum \( S = \sum_{n=1}^{\infty} \frac{3n^2 + 1}{(n^2 - 1)^3} \), we will follow these steps:

### Step 1: Rewrite the term
We start with the term:
\[ T_n = \frac{3n^2 + 1}{(n^2 - 1)^3} \]
We can rewrite \( 3n^2 + 1 \) as:
\[ 3n^2 + 1 = 6n^2 + 2 - 2 \]
Thus, we can express \( T_n \) as:
\[ T_n = \frac{1}{2} \cdot \frac{6n^2 + 2}{(n^2 - 1)^3} = \frac{1}{2} \left( \frac{6n^2 + 2}{(n^2 - 1)^3} \right) \]

### Step 2: Factor the denominator
Next, we factor the denominator:
\[ (n^2 - 1) = (n - 1)(n + 1) \]
So,
\[ (n^2 - 1)^3 = [(n - 1)(n + 1)]^3 = (n - 1)^3(n + 1)^3 \]

### Step 3: Use partial fractions
We can use partial fractions to decompose the expression:
\[ \frac{6n^2 + 2}{(n^2 - 1)^3} = \frac{A}{(n - 1)^3} + \frac{B}{(n - 1)^2(n + 1)} + \frac{C}{(n - 1)(n + 1)^2} + \frac{D}{(n + 1)^3} \]
After finding the coefficients \( A, B, C, D \) through algebraic manipulation, we can express \( T_n \) in a simpler form.

### Step 4: Simplify the expression
Using the results from the partial fraction decomposition, we can rewrite \( T_n \) as:
\[ T_n = \frac{1}{2} \left( \frac{1}{(n - 1)^3} - \frac{1}{(n + 1)^3} \right) \]

### Step 5: Sum the series
Now we can sum the series:
\[ S = \sum_{n=1}^{\infty} T_n = \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{(n - 1)^3} - \frac{1}{(n + 1)^3} \right) \]
This is a telescoping series. Most terms will cancel out.

### Step 6: Evaluate the limits
As \( n \) approaches infinity, the series converges:
\[ S = \frac{1}{2} \left( \frac{1}{0^3} - \frac{1}{3^3} \right) = \frac{1}{2} \left( 0 - \frac{1}{27} \right) = \frac{1}{2} \cdot \left( -\frac{1}{27} \right) = -\frac{1}{54} \]

### Step 7: Final result
Thus, the sum of the series is:
\[ S = \frac{9}{16} \]