# How do you find the 4-th partial sum of the infinite series #sum_(n=1)^oo(3/2)^n# ?

To find the fourth partial sum of the infinite series (\sum_{n=1}^{\infty} (3/2)^n), use the formula for the partial sum of a geometric series:

[S_n = a \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. In this series, (a = \frac{3}{2}) and (r = \frac{3}{2}).

[S_4 = \frac{3}{2} \frac{1 - (\frac{3}{2})^4}{1 - \frac{3}{2}}]

[S_4 = \frac{3}{2} \frac{1 - \frac{81}{16}}{1 - \frac{3}{2}}]

[S_4 = \frac{3}{2} \frac{\frac{16}{16} - \frac{81}{16}}{\frac{2}{2} - \frac{3}{2}}]

[S_4 = \frac{3}{2} \frac{-\frac{65}{16}}{-\frac{1}{2}}]

[S_4 = \frac{3}{2} \times \frac{65}{16} \times 2]

[S_4 = \frac{195}{16}]

So, the fourth partial sum of the series is (\frac{195}{16}).

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This is geometric progression series of which

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When dealing with a sum, you have a sequence that generates the terms. In this case, you have the sequence

You may compute each term, but there is a useful formula:

So, in your case

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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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