Trouble finding a general formula for the nth term

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SUMMARY

The discussion focuses on deriving a general formula for the nth term of the sequence {1, 1/2, -1/6, -7/24, 1/24, 7/48, -19/336, -503/4480, 17257/362880, 12913/145152,...}. The user suggests that factorials (n!) may play a crucial role in the denominators of the terms. The proposed method involves arranging terms to have a common denominator of n!, factoring numerators into primes, and graphically analyzing the presence of primes to identify patterns. The approach emphasizes iterative refinement of the formula based on observed patterns in the numerators.

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Im having trouble finding a general formula for the nth term for the sequence

{1, 1/2, -1/6, -7/24, 1/24, 7/48, -19/336, -503/4480, 17257/362880,

12913/145152,...}

if one exist

so far I am guessing that n! has something to do with it

a hint would be super cool!
 
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Maybe you could tell us where you got this. If it's a homework problem given to you as is, why do you wonder if a pattern exists? In any case, the n! in the denominator might not be a result of an explicit n! in the pattern, it might be that the pattern is produced by requiring that the sum of some combination of the first n terms is some function of n, the n! arising when you add the fractions using a common denominator.
 
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Take your terms and, like you think, arrange them so that the denominators are all equal to n!. This will make some of your numerators larger.

Second stage is to go and factor each numerator into primes. If this method is to have a prayer of a chance, even the last (largest) numerator should factor so that its largest prime factor is still relatively small, maybe less than 100.

Third stage is to take a piece of graph paper and mark n along the top (you've got n=1 through n=10 data above). Mark the odd integers down the the horizontal direction, hopefully from 1 until you reach the largest prime factor that was included in the numerators.

Now put an x in the array anywhere a prime appears. Look for a pattern.

A pattern would be something like several of the numbers happening to have primes in the numerator that are in a straight line. These lines will be missing elements, but those elements are likely composite odd numbers that can be obtained by multiplying primes in the missing n values.

For any line that looks like its good (the steepest lines are the most important), figure out the equation for that line and then use that to modify the original problem. For example, if you find that (7n+5) appears, you might try taking your series and dividing it by (7n+5). Then go back and repeat finding the prime factors.

Each time you repeat this, the theory is that you're going to pull factors out of the formula. Eventually, you may get the answer.

Of course all this might not work, but it's the way I'd approach the problem (in general).

And you might look at what kind of hints the book and the instructor (if this is for a class) has been throwing around.

Carl
 

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