# 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*3)+...+1/(n(n+1))+...#?

To find the nth partial sum of the series ( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \ldots + \frac{1}{n(n+1)} ), we first need to express each term in a more general form. Notice that ( \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} ).

So, the nth partial sum ( S_n ) can be written as:

[ S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{n} - \frac{1}{n+1} \right) ]

Notice that most of the terms will cancel out, leaving only ( \frac{1}{1} - \frac{1}{n+1} ). Simplifying, we get:

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

To determine whether the series converges, we need to take the limit as ( n ) approaches infinity of the nth partial sum.

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

Since the limit exists and is finite, the series converges.

To find the sum when it exists, we use the value of the limit:

[ \text{Sum} = 1 ]

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Given the series:

we can determine that is convergent using the direct comparison test since:

and the series:

Decompose now the general term using partial fractions:

and note that in the partial sum:

all terms cancel each other except the first and the last, so:

and then:

is the sum of the series.

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The answer is

As

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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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