sum_(n=-1)^infinity(-1)^(n+1) (1/3)^n = 9/4~~2.25... How do you prove that? Is there a certain formula to figure out the sum of the alternating series?
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Originally Posted by Brotha sum_(n=-1)^infinity(-1)^(n+1) (1/3)^n = 9/4~~2.25... How do you prove that? Is there a certain formula to figure out the sum of the alternating series? It appears to be geometric in form. Observe that: $\displaystyle \sum_{n=-1}^{\infty}(-1)^{n+1}\left(\tfrac{1}{3}\right)^n=3-\sum_{n=0}^{\infty}\left(-\tfrac{1}{3}\right)^n=\ldots$ Can you finish this?
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