# If S = sigma from n=2 to infinity of (3n^2 + 1) divided by (n^2 - 1) cubed, then 16S is:

## Question

If $S=\sum_{n=2}^{\infty} \frac{3n^2+1}{(n^2-1)^3}$ then $16S$ is:

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## Answer

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Given:

$$ S = \sum_{n=2}^{\infty} \frac{3n^2 + 1}{(n^2 - 1)^3} $$

We start by recognizing that:

$$ (n^2 - 1)^3 = (n-1)^3 (n+1)^3 $$

Additionally:

$$ (n+1)^3 - (n-1)^3 = 6n^2 + 2 $$

Now, we can express the series as follows:

$$ S = \sum_{n=2}^{\infty} \frac{1}{2} \left(\frac{(n+1)^3 - (n-1)^3}{(n+1)^3 (n-1)^3}\right) $$

Rewriting the sum, we get:

$$ S = \frac{1}{2} \sum_{n=2}^{\infty} \left( \frac{1}{(n-1)^3} - \frac{1}{(n+1)^3} \right) $$

This can be expanded as a telescoping series:

$$ S = \frac{1}{2} \left( \left( \frac{1}{1^3} - \frac{1}{3^3} \right) + \left( \frac{1}{2^3} - \frac{1}{4^3} \right) + \left( \frac{1}{3^3} - \frac{1}{5^3} \right) + \ldots \right) $$

The terms $\frac{1}{3^3}, \frac{1}{4^3}, \ldots$ cancel out in the series, leaving us with:

$$ S = \frac{1}{2} \left( \frac{1}{1^3} + \frac{1}{2^3} \right) $$

Simplifying further:

$$ S = \frac{1}{2} \left( 1 + \frac{1}{8} \right) = \frac{1}{2} \left( \frac{9}{8} \right) = \frac{9}{16} $$

Thus,

$$ 16S = 16 \cdot \frac{9}{16} = 9 $$

Therefore, the value of $16S$ is:

$$ \boxed{9} $$

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