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August 7th, 2007, 06:49 PM  #1 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  addition of fractions
1  1/2 + 1/3  1/4 + ...  1/1318 + 1/1319?

August 8th, 2007, 04:45 AM  #2 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
Are they assuming an infinite alternating series, or just the specified number of terms?

August 8th, 2007, 05:04 AM  #3 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
it's not infinite  it's specified number of terms

August 8th, 2007, 07:47 AM  #4 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
Golly, that could be pretty hard. You might be able to simplify things by splitting the sequence into two separate sequences, one with all the positive terms, and the other with all the negative terms. You can then add them back together when you figure out their individual sums.

August 8th, 2007, 09:08 AM  #5  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: addition of fractions Quote:
97124353701764572117890321122757050354360580230566 6817216269112276964353613 49633888874185035182122277071818873992844213242099 915802613721808810660422673400 13822543021630993790301871917306672479003742347731 906894502206611684973377349822 14072786674879347231455514608195413335921969322484 234571620156283811311048866422 99904432585085953323951296172861435808985669081282 324361874663718716410956082578 59832669859587159077972843255741143946005144546720 084104126426401254209483858435 52514663707886945205678870894460896001671218040951 566830190968605636365403982335 1264466923197720153813/14004426370194880461934343682044774391282833031127 9107462 97786781959105584828861694180270900155320736866485 178357969757814225700469374818 45418627056015641968611654202902769674287048827771 357301459887020639751248870614 15171934261220862316341780802377230654486825981311 825976235273131790688782312984 72650458406330723014358986732179004322578669107971 003162372262774921109835980551 50076789879745230479962807184918663328472109316343 903768300078189869559464609212 48679410595324531407209849932757037376488884192237 742369637229672994844934665387 92766661068428697404447683623975519360000 You can approximate this yourself by taking the sum in pairs: (1  1/2) + (1/3  1/4) + ....  
August 8th, 2007, 10:44 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,931 Thanks: 2205 
Really?

August 8th, 2007, 11:39 AM  #7 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
I would be surprised if every digit after that were a zero, which is implicit in the claim that that is an exact answer. Any way, the question specifies that it stop at the 1319th term, rather than the limit as the term number approaches infinity.

August 8th, 2007, 12:44 PM  #8 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
My guess is that that huge decimal number has something to do with 1319!, the common denominator of all the fractions.

August 8th, 2007, 12:58 PM  #9  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
Quote:
And yes, it's exact.  
August 8th, 2007, 06:22 PM  #10 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
Okay, I see the slash. At first I assumed that that was a continuation of the decimal you already gave. Followup question: does the series converge, and, if so, to what?


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