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

To find the nth partial sum of the series (1 - 2 + 3 - 4 + \ldots + n(-1)^{n-1}), you need to analyze the pattern. The series alternates between positive and negative terms based on the power of (-1), with each term being the consecutive integer multiplied by (-1) raised to the power of one less than its position.

The nth partial sum (S_n) can be calculated using this pattern. If n is even, the sum will be negative, and if n is odd, the sum will be positive.

To determine whether the series converges, you can check if the sequence of nth partial sums approaches a limit as n approaches infinity. In this case, the series diverges because the terms do not approach a specific value but rather alternate between positive and negative integers.

Therefore, the series does not converge, and there is no finite sum when it exists.

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is indeterminate.

The series does not converge as it does not satisfy the necessary condition for convergence:

Hence the sums oscillate and are unbounded in absolute value so that:

does not exist.

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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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