# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #(1-1/2)+(1/2-1/6)+(1/6-1/24)+...+(1/(n!)-1/((n+1)!))+...#?

To find the nth partial sum of the series ( (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{6}) + (\frac{1}{6} - \frac{1}{24}) + \ldots + (\frac{1}{n!} - \frac{1}{(n+1)!}) + \ldots ), you can observe that each term can be expressed as the difference between two consecutive terms in the sequence ( \frac{1}{n!} ). Thus, the nth partial sum can be written as:

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

To determine whether the series converges, you can take the limit of the nth partial sum as ( n ) approaches infinity:

[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(1 - \frac{1}{(n+1)!}\right) = 1 - \lim_{n \to \infty} \frac{1}{(n+1)!} = 1 ]

Since the limit of the nth partial sum exists and is finite, the series converges.

To find the sum of the series when it exists, you use the fact that the limit of the nth partial sum approaches the sum of the series as ( n ) approaches infinity. Therefore, the sum of the series is:

[ \sum_{n=1}^{\infty} \left(1 - \frac{1}{(n+1)!}\right) = 1 ]

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The

and the series converges to

A series like this, in which all but a finite number of terms cancel out, is called a telescoping series.

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The series converges to unity.

Note that almost all the terms cancel, leaving:

And so,

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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