document.write( "Question 887117: I need to find the sum of the series (1/2^n) from 1 approaching infinity.\r
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\n" ); document.write( "\n" ); document.write( "my first term is 1/2 and Im having difficulty finding my ratio. I first thought my ratio was 2 since that is what n is being multiplied by. But that didn't work. I also tried 1/2 but that didn't work either.
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Algebra.Com's Answer #536376 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "\"1%2F2\" does work!\r\n" );
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document.write( "You are talking about \"%22%22=%22%22\"\"1%2F2%5E1%2B1%2F2%5E2%2B1%2F2%5E3%2B%22%22%2A%22%22%2A%22%22%2A%22%22\"\r\n" );
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document.write( "Certainly the common ratio is \"r=1%2F2\" because any term after the first\r\n" );
document.write( "divided by the preceding term is always \"1%2F2\"\r\n" );
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document.write( "\"S%5Binfinity%5D\"\"%22%22=%22%22\"\"a%5B1%5D%2F%281-r%29\", with \"a%5B1%5D=1%2F2\", \"r=1%2F2\"\r\n" );
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document.write( "\"S%5Binfinity%5D\"\"%22%22=%22%22\"\"%281%2F2%29%2F%281-expr%281%2F2%29%29\"\"%22%22=%22%22\"\"%281%2F2%29%2F%281%2F2%29\"\"%22%22=%22%22\"\"1\".\r\n" );
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document.write( "Edwin
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