document.write( "Question 537595: EVALUATE:\r
\n" ); document.write( "\n" ); document.write( "\"1%2B3%2F4%2B7%2F16%2B15%2F64%2B31%2F256\" ......... to infinity
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Algebra.Com's Answer #352942 by AnlytcPhil(1806)\"\" \"About 
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\"1%2B3%2F4%2B7%2F16%2B15%2F64%2B31%2F256\" ......... to infinity
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\n" ); document.write( "Consider the first term as \"1%2F1\"
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\n" ); document.write( "\"1%2F1%2B3%2F4%2B7%2F16%2B15%2F64%2B31%2F256\" ......... to infinity
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\n" ); document.write( "The numerators form the sequence 1,3,7,15,31
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\n" ); document.write( "That has nth term
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\n" ); document.write( "\"2%5En-1\"
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\n" ); document.write( "The denominators form the sequence 1,4,16,64,256
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\n" ); document.write( "That has nth term
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\n" ); document.write( "\"2%5E%28n-1%29\"
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\n" ); document.write( "Therefore the sequence summed to infinity is
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\n" ); document.write( "\"sum%28++%28%282%5En-1%29%2F%282%5E%28n-1%29%29%29%2Cn=1%2Cinfinity+%29\" = \"sum%28++%28%282%5En%29%2F%282%5E%28n-1%29%29%29%2Cn=1%2Cinfinity+%29\" - \"sum%28++%281%2F%282%5E%28n-1%29%29%29%2Cn=1%2Cinfinity+%29\" =
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\n" ); document.write( "Subtract exponents in the first:
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\n" ); document.write( "\"sum%28++%282%5E1%29%2Cn=1%2Cinfinity+%29\" - \"sum%28++%281%2F%282%5E%28n-1%29%29%29%2Cn=1%2Cinfinity+%29\"
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\n" ); document.write( "The first series diverges to infinity and the second one is a geometric
\n" ); document.write( "series converging to 2, so the series diverges to infinity.
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\n" ); document.write( "Edwin
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