# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #ln(1/2)+ln(2/3)+ln(3/4)+...+ln(n/(n+1))+...#?

To find the nth partial sum of the series ( \ln\left(\frac{n}{n+1}\right) ), we need to sum the terms from ( n = 1 ) to ( n = N ).

The nth term is ( \ln\left(\frac{n}{n+1}\right) ).

The nth partial sum, denoted as ( S_N ), is obtained by summing the first ( N ) terms of the series.

To determine whether the series converges, we can investigate its behavior as ( N ) approaches infinity. If the series approaches a finite value as ( N ) goes to infinity, then it converges. Otherwise, it diverges.

To find the sum when it exists, we take the limit of the nth partial sum as ( N ) approaches infinity.

[ S = \lim_{N \to \infty} S_N ]

[ S = \lim_{N \to \infty} \sum_{n=1}^{N} \ln\left(\frac{n}{n+1}\right) ]

[ S = \lim_{N \to \infty} \ln\left(\frac{1}{2}\right) + \ln\left(\frac{2}{3}\right) + \ln\left(\frac{3}{4}\right) + \ldots + \ln\left(\frac{N}{N+1}\right) ]

[ S = \lim_{N \to \infty} \ln\left(\frac{1}{N+1}\right) ]

[ S = \ln\left(\lim_{N \to \infty} \frac{1}{N+1}\right) ]

[ S = \ln(0) ]

As ( N ) approaches infinity, the term ( \frac{1}{N+1} ) approaches 0, which makes the natural logarithm of 0 undefined.

Thus, the series diverges.

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So, the sum of the infinite series diverges.

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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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