Guest Posted April 27, 2008 Report Share Posted April 27, 2008 Compute the sum till infinity S = 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1) +....................... Quote Link to comment Share on other sites More sharing options...
0 grey cells Posted April 27, 2008 Report Share Posted April 27, 2008 Compute the sum till infinity S = 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1) +....................... Infinity. or is it 0? Anything + infinity=infinity. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 27, 2008 Report Share Posted April 27, 2008 (edited) Infinity. or is it 0? Anything + infinity=infinity. If it was zero, it wouldn't have posted it. And I didn't get the logic behind infinity. How can you say that the sum is infinite? Edited April 27, 2008 by esjohnson Quote Link to comment Share on other sites More sharing options...
0 grey cells Posted April 27, 2008 Report Share Posted April 27, 2008 I didn't get the logic behind infinity. How can you say that the sum is infinite? you had specified in the question that , compute till infinity , I had arrived at the answer infinity. Basically , till infinity means , any series should continue till infinity . Quoting your sequence: S = 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1) +.......................+infinity This would ensure that the sum is infinity.As you had not specified a finite value , the sequence must continue till infinity. But if the sequence continued with alternating 1's and (-1)'s , the result could be -1 or 0. Please correct me if I am wrong. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 27, 2008 Report Share Posted April 27, 2008 you had specified in the question that , compute till infinity , I had arrived at the answer infinity. Basically , till infinity means , any series should continue till infinity . Quoting your sequence: S = 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1) +.......................+infinity This would ensure that the sum is infinity.As you had not specified a finite value , the sequence must continue till infinity. But if the sequence continued with alternating 1's and (-1)'s , the result could be -1 or 0. Please correct me if I am wrong. No no, by 'till infinity ' I meant that its an infinite sequence . And, The answer is neither -1 nor 0 . Quote Link to comment Share on other sites More sharing options...
0 grey cells Posted April 27, 2008 Report Share Posted April 27, 2008 No no, by 'till infinity ' I meant that its an infinite sequence . And, The answer is neither -1 nor 0 . Sorry , I misunderstood it . But not my fault , because many puzzles are posted , with lines meant to distract the readers , i;e. , googlies. 1? I have realized that -1 is not a possibility. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 27, 2008 Report Share Posted April 27, 2008 Sorry , I misunderstood it . But not my fault , because many puzzles are posted , with lines meant to distract the readers , i;e. , googlies. 1? I have realized that -1 is not a possibility. Sorry, the answer is not 1 . Quote Link to comment Share on other sites More sharing options...
0 grey cells Posted April 27, 2008 Report Share Posted April 27, 2008 (-1)^n. Am I anywhere close? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 27, 2008 Report Share Posted April 27, 2008 (-1)^n. Am I anywhere close? If it was (-1)n, then the answers -1,1 for n=1,2 would have been correct . hence (-1)n is not the answer too . No, you are not close anywhere . Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 27, 2008 Report Share Posted April 27, 2008 (-1)^n. Am I anywhere close? If it was (-1)n, then the answers -1,1 for n=1,2 would have been correct . hence (-1)n is not the answer too . No, you are not close anywhere . Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 27, 2008 Report Share Posted April 27, 2008 would it beS=S Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 27, 2008 Report Share Posted April 27, 2008 would it beS=S Wow ,now I can't say no as thats an universal answer . But, what is S ? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 27, 2008 Report Share Posted April 27, 2008 you cant do it can u? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 27, 2008 Report Share Posted April 27, 2008 how about .... S=1/2. S=1+(-1)+1+(-1)+1 ... S= 1+(-1)+1+(-1)+ ... Adding both equations 2S=1 ==> S=1/2. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 27, 2008 Report Share Posted April 27, 2008 how about .... S=1/2. S=1+(-1)+1+(-1)+1 ... S= 1+(-1)+1+(-1)+ ... Adding both equations 2S=1 ==> S=1/2. Nice try , but thats not right . Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted April 28, 2008 Report Share Posted April 28, 2008 Depending on how you group the terms it could be either 1 or 0. [1] If you group each odd term with the following even term, you are adding a succession of 0's, and the sum is zero. [2] If you leave the 1st term alone and group the 2nd and succeeding even terms with the following odd terms, you start with 1 and add an infinite number of 0's to it, and the sum is 1. [3] But since the partial sums alternate between 1 and 0, it's more satisfying to my mind to say the sum is undetermined. There is no value to which the partial sums become arbitrarily close as the number of terms increases without bound. So take your choice of answers: [1] 0. For the glass is half empty folk [2] 1. For the glass is half full people [3] undefined. For the mathematicians. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 28, 2008 Report Share Posted April 28, 2008 (edited) I'm going to use 8 to stand for infinity... S= 8 (1) + 8 (-1) S= 8 [1+(-1)] S= 8 (0) it still looks like it should be 0 but maybe it's just infinity? Edited April 28, 2008 by krazychik Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 28, 2008 Report Share Posted April 28, 2008 Depending on how you group the terms it could be either 1 or 0. [1] If you group each odd term with the following even term, you are adding a succession of 0's, and the sum is zero. [2] If you leave the 1st term alone and group the 2nd and succeeding even terms with the following odd terms, you start with 1 and add an infinite number of 0's to it, and the sum is 1. [3] But since the partial sums alternate between 1 and 0, it's more satisfying to my mind to say the sum is undetermined. There is no value to which the partial sums become arbitrarily close as the number of terms increases without bound. So take your choice of answers: [1] 0. For the glass is half empty folk [2] 1. For the glass is half full people [3] undefined. For the mathematicians. Yes, bonanova got it. As we can see from the previous post that the answer could be 0,1,1/2....etc Therefore the answer is indeterminate Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 28, 2008 Report Share Posted April 28, 2008 you cant do it can u? Brain Master was on right track too . Quote Link to comment Share on other sites More sharing options...
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Compute the sum till infinity
S = 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1) +.......................
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