# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1-1/3+1/9-...+(-1/3)^n+...#?

To find the nth partial sum of the series (1 - \frac{1}{3} + \frac{1}{9} - \ldots + (-1/3)^n + \ldots), you use the formula for the partial sum of a geometric series:

[S_n = a \cdot \frac{1 - r^n}{1 - r}]

where (a) is the first term of the series, (r) is the common ratio, and (n) is the number of terms.

For this series, (a = 1) and (r = -\frac{1}{3}). So, the nth partial sum is:

[S_n = 1 \cdot \frac{1 - (-1/3)^n}{1 - (-1/3)}]

To determine whether the series converges, you need to examine the common ratio (r). Since (|r| < 1), the series converges.

To find the sum when it exists, use the formula for the sum of an infinite geometric series:

[S = \frac{a}{1 - r}]

where (a) is the first term and (r) is the common ratio.

For this series, (a = 1) and (r = -\frac{1}{3}). So, the sum of the series is:

[S = \frac{1}{1 - (-1/3)}]

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The Maclaurin series

Here, x = 1/3.

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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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