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could any one give me hint how to solve $$\sum_{n=1}^\infty \frac{3}{(-2)^n} $$

using the geometric series?

DSL
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1 Answers1

1

Write

$$ 3 \sum_{n \geq 1} \left( \frac{-1}{2} \right)^n = 3 \sum_{n \geq 0} \left( \frac{-1}{2} \right)^n - 3 = 3 \cdot \frac{ 1 }{1 + \frac{1}{2}} -3 = 3 \cdot \frac{2}{3} - 3 = -1$$

ILoveMath
  • 10,694
  • @ ILoveMath should it be -1? – DSL Nov 07 '16 at 05:36
  • you are right, just fixed it – ILoveMath Nov 07 '16 at 05:39
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    The sum is $S = 3\cdot \frac{a}{1-r} = 3\cdot \frac{-\frac 12}{1 - (-\frac 12)} = -3\cdot \frac 13 = -1$ Oh, you just fixed it, never mind. – Deepak Nov 07 '16 at 05:39
  • @ ILoveMath I just figured how to use the geometric series to find the limit as n approaches infinity, although your answer did not strictly used the geometric series, it did help me. thanks. – DSL Nov 07 '16 at 05:46