$$ \sum _ { k = 1 } ^ { \infty } \frac { 1 } { ( k + 1 ) ( k - 1 ) ! } , \quad S _ { n } = 1 - \frac { 1 } { ( n + 1 ) ! } $$

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UpStudy AI · Accepted Answer

The limit of $$ S_n $$ as $$ n $$ approaches infinity is not directly calculable using the given expression.

However, we can observe that the series is a telescoping series, where each term cancels out the previous term.

Let's rewrite the series to see the telescoping pattern:

$$ \sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!} = \sum_{k=1}^{\infty} \left( \frac{1}{k!} - \frac{1}{(k+1)!} \right) $$

This series can be simplified by canceling out the terms:

$$ \sum_{k=1}^{\infty} \left( \frac{1}{k!} - \frac{1}{(k+1)!} \right) = \left( \frac{1}{1!} - \frac{1}{2!} \right) + \left( \frac{1}{2!} - \frac{1}{3!} \right) + \left( \frac{1}{3!} - \frac{1}{4!} \right) + \ldots $$

The terms cancel out, leaving only the first term $$ \frac{1}{1!} $$.

Therefore, the sum of the series is $$ \frac{1}{1!} = 1 $$.
The sum of the series is 1.

\( \sum _ { k = 1 } ^ { \infty } \frac { 1 } { ( k + 1 ) ( k - 1 ) ! } , \quad S _ { n } = 1 - \frac { 1 } { ( n + 1 ) ! } \)

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