document.write( "Question 1010159: How do you solve 1/2+2/3+3/4+4/5......+2013/2014? \n" ); document.write( "
Algebra.Com's Answer #625654 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
1/2+2/3+3/4+4/5......+2013/2014 =
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document.write( "That has 2013 terms, and is equal to\r\n" );
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document.write( "(1-1/2)+(1-1/3)+(1-1/4)+(1-1/5)+...+(1-1/2014) =\r\n" );
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document.write( "(1+1+...+1)-1/2-1/3-1/4-1/5-...-1/2014) =\r\n" );
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document.write( "There are 2013 1's added so this is:\r\n" );
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document.write( "2013 - (1/2+1/3+1/4+1/5+...+1/2014)\r\n" );
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document.write( "That series in parentheses starts with 1/2.\r\n" );
document.write( "The harmonic series starts with 1/1, so let's\r\n" );
document.write( "add and subtract 1:\r\n" );
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document.write( "which equals:\r\n" );
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document.write( "2013 + 1 - 1 - (1/2+1/3+1/4+1/5+...+1/2014) =\r\n" );
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document.write( "2014 - (1+1/2+1/3+1/4+1/5+...+1/2014)\r\n" );
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document.write( "In the parentheses in the harmonic sequence to 2014 terms.\r\n" );
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document.write( "There is no formula for that, but a good approximation for it is:\r\n" );
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document.write( "Hn = ln(n) + g + 1/(2n) - 1/(12n^2)\r\n" );
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document.write( "where g = the Euler-Mascheroni constant, which\r\n" );
document.write( "is approximately 0.5772156649.  You can Google information on that.\r\n" );
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document.write( "Substituting n = 2014\r\n" );
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document.write( "H2014 is approximately 8.185342021, making the desired sum\r\n" );
document.write( "approximately\r\n" );
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document.write( "2014 -  8.185342021 = 2005.814658\r\n" );
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document.write( "When I ran that sum on the computer, I got 2005.81466\r\n" );
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document.write( "Edwin
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