# How do you find the 10-th partial sum of the infinite series #sum_(n=1)^oo(0.6)^(n-1)# ?

To find the 10th partial sum of the infinite series (\sum_{n=1}^{\infty} (0.6)^{n-1}), you can use the formula for the sum of a geometric series:

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

Where: (S_n) = the sum of the first (n) terms (a) = the first term of the series (r) = the common ratio (n) = the number of terms

In this series, (a = 1) (the first term is (0.6^0 = 1)) and (r = 0.6).

Plugging these values into the formula, you get:

[S_{10} = \frac{1(1 - (0.6)^{10})}{1 - 0.6}]

[S_{10} = \frac{1 - 0.6^{10}}{1 - 0.6}]

Now you can calculate (S_{10}) using a calculator or computational tool.

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As an extra brain rep, see if you can verify this formula using our subtraction method, and then maybe figure out the infinite sum.

\dansmath - made from scratch, every time./

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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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