How do you find the sum of the infinite geometric series 4+2+0-2...?

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How do you find the sum of the infinite geometric series 4+2+0-2...?

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Answer 1 · 2015-11-14T08:38:05

This is an arithmetic series, not a geometric one.

Its sum diverges to $- \infty$ $-oo$ .

Explanation:

The terms of this series are of the form $a_{n} = a_{1} + (n - 1) d$ $a_n = a_1 + (n-1)d$ , with initial term $a_{1} = 4$ $a_1 = 4$ and common difference $d = - 2$ $d = -2$

$N \sum n = 1 a_{n} = N \sum n = 1 (a_{1} + (n - 1) d) = N (a_{1} + \frac{(N - 1) d}{2})$ $sum_(n=1)^N a_n = sum_(n=1)^N (a_1 + (n-1)d) = N(a_1 + ((N-1)d)/2)$

$= 4 N + \frac{N (N - 1) \cdot (- 2)}{2} = 4 N - N (N - 1)$ $=4N + (N(N-1)*(-2))/2 = 4N-N(N-1)$

$= 5 N - N^{2} \to - \infty$ $= 5N-N^2 -> -oo$ as $N \to \infty$ $N->oo$

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How do you find the sum of the infinite geometric series 4+2+0-2...?

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