document.write( "Question 146982This question is from textbook engineering mathematics
\n" ); document.write( ": please help me solve this sum on summation of series:\r
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Algebra.Com's Answer #107964 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
please help me solve this sum on summation of series:
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\n" ); document.write( "\"1%2F%281%2A2%2A3%29+%2B+3%2F%282%2A3%2A4%29+%2B+5%2F%283%2A4%2A5%29+%2B+7%2F%284%2A5%2A6%29\" + ···
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document.write( "The numerators are the odd numbers, and the nth odd integer\r\n" );
document.write( "is 2n-1, so that will go on top in the nth term formula.  The \r\n" );
document.write( "denominator of the nth term formula is the product of n and \r\n" );
document.write( "the next two integers, (n+1) and (n+2), so the nth term is\r\n" );
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document.write( "\"%282n-1%29%2F%28n%28n%2B1%29%28n%2B2%29%29\" \r\n" );
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document.write( "\"1%2F%281%2A2%2A3%29+%2B+3%2F%282%2A3%2A4%29+%2B+5%2F%283%2A4%2A5%29+%2B+7%2F%284%2A5%2A6%29\" + ··· = \"sum%28%282n-1%29%2F%28n%28n%2B1%29%28n%2B2%29%29%2Cn=1%2C+infinity%29\" \r\n" );
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document.write( "Do you understand partial fractions?  If not post again\r\n" );
document.write( "asking how.  But I'll assume you already know how.\r\n" );
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document.write( "We break the summand into partial fractions:\r\n" );
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document.write( "Now suppose we sum this just to some large positive integer K,\r\n" );
document.write( "and then take the limit as K approaches infinity:\r\n" );
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document.write( "lim   \r\n" );
document.write( "K->oo\r\n" );
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document.write( "Make three sums:\r\n" );
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document.write( "lim   \r\n" );
document.write( "K->oo\r\n" );
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document.write( "Put the constant multipliers in front of the summations:\r\n" );
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document.write( "lim   \r\n" );
document.write( "K->oo\r\n" );
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document.write( "Make the denominators in the second and third summands all be single\r\n" );
document.write( "letters, by setting them equal to another letter:\r\n" );
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document.write( "In the second summation, let \"n%2B1=M\" or \"n=M-1\" and\r\n" );
document.write( "in the third summation, let \"n%2B2=P\" or \"n=P-2\"\r\n" );
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document.write( "lim   \r\n" );
document.write( "K->oo\r\n" );
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document.write( "lim   \r\n" );
document.write( "K->oo\r\n" );
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document.write( "Now no change will result if we replace M and P by n:\r\n" );
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document.write( "lim   \r\n" );
document.write( "K->oo\r\n" );
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document.write( "In the first summation we write out the first two terms, and\r\n" );
document.write( "write the summation from n=3 to n=K\r\n" );
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document.write( "In the second summation we write out the first term and the last\r\n" );
document.write( "term and write the summation from n=3 to n=K\r\n" );
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document.write( "In the third summation we write out the last two\r\n" );
document.write( "terms and write the summation from n=3 to n=K\r\n" );
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document.write( "Now we let \"sum%28%281%2Fn%29%2Cn=3%2Cn=K%29=S\"\r\n" );
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document.write( "So, now the sums become:\r\n" );
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document.write( "\"sum%28%281%2Fn%29%2Cn=1%2Cn=K%29=3%2F2%2BS\"\r\n" );
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document.write( "\"sum%28%281%2Fn%29%2Cn=2%2Cn=K%2B1%29=1%2F2%2BS%2B1%2F%28K%2B1%29\"\r\n" );
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document.write( "\"sum%28%281%2Fn%29%2Cn=3%2Cn=K%2B2%29=S%2B1%2F%28K%2B1%29%2B1%2F%28K%2B2%29%29\"\r\n" );
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document.write( "So,\r\n" );
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document.write( "lim   \r\n" );
document.write( "K->oo\r\n" );
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document.write( "becomes:\r\n" );
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document.write( "lim   \r\n" );
document.write( "K->oo\r\n" );
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document.write( "or:\r\n" );
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document.write( "lim   \r\n" );
document.write( "K->oo\r\n" );
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document.write( "Write the \"3S\" as \"6S%2F2\" and \"3%2F2\" as \"6%2F4\"\r\n" );
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document.write( "lim   \r\n" );
document.write( "K->oo\r\n" );
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document.write( "Ths terms in S all cancel out, and we have\r\n" );
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document.write( "lim   \"%283%2F4%2B3%2F%28K%2B2%29-5%2F%282%28K%2B1%29%29-5%2F%282%28K%2B3%29%29%29%29\"\r\n" );
document.write( "K->oo\r\n" );
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document.write( "The last three terms approach 0 as K approaches infinity.\r\n" );
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document.write( "Therefore the answer is \"3%2F4\"\r\n" );
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document.write( "Edwin
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