document.write( "Question 1184837: Please help me answer this question --\r
\n" ); document.write( "\n" ); document.write( "Test whether the following infinite series converge, and if they do, find their sum:\r
\n" ); document.write( "\n" ); document.write( "i) 1/(1*2) - 1/(2*3) + 1/(3*4) - 1/(4*5) + 1/(5*6) - ...\r
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\n" ); document.write( "\n" ); document.write( "ii) 1/(1*2) - 1/(3*4) + 1/(5*6) - 1/(7*8) + 1/(9*10) - ...\r
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Algebra.Com's Answer #815493 by ikleyn(52798)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "                    I will answer part  (i),   ONLY.\r
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\n" ); document.write( "\n" ); document.write( "(i)   1/(1*2) - 1/(2*3) + 1/(3*4) - 1/(4*5) + 1/(5*6) - ...
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document.write( "\"1%2F%281%2A2%29\" = \"1\" - \"1%2F2\"\r\n" );
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document.write( "\"1%2F%282%2A3%29\" = \"1%2F2\" - \"1%2F3\"\r\n" );
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document.write( "\"1%2F%283%2A4%29\" = \"1%2F3\" - \"1%2F4\"\r\n" );
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document.write( "\"1%2F%284%2A5%29\" = \"1%2F4\" - \"1%2F5\"\r\n" );
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document.write( "\"1%2F%285%2A6%29\" = \"1%2F5\" - \"1%2F6\"\r\n" );
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document.write( "  . . . . . . . . . . . . \r\n" );
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document.write( "\"1%2F%28k%2A%28k%2B1%29%29\" = \"1%2Fk\" - \"1%2F%28k%2B1%29\"\r\n" );
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document.write( "Now take the alternate sum and get\r\n" );
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document.write( "    S = (1*2) - 1/(2*3) + 1/(3*4) - 1/(4*5) + 1/(5*6) - . . . = (\"1\" - \"1%2F2\") - (\"1%2F2\" - \"1%2F3\") + (\"1%2F3\" - \"1%2F4\") - (\"1%2F4\" - \"1%2F5\") + (\"1%2F5\" - \"1%2F6\" ) + . . . = \r\n" );
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document.write( "      = 1 - 2/2 + 2/3 - 2/4 + 2/5 - 2/6 + . . . = (2 - 2/2 + 2/3 - 2/4 + 2/5 - 2/6  + . . . ) - 1 = 2*(1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + . . . ) - 1 = \r\n" );
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document.write( "      = 2*ln(1-(-1)) - 1 = 2*ln(2) - 1 = 0.386294361  (rounded).    ANSWER\r\n" );
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\n" ); document.write( "\n" ); document.write( "For more details,  see this link \r
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\n" ); document.write( "\n" ); document.write( "https://www.quora.com/What-is-1-1-2+1-3-1-4+1-5-1-6\r
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