Split Summation
True or False? n = 1 ∑ ∞ n ( n + 1 ) 1 = n = 1 ∑ ∞ n 1 − n = 1 ∑ ∞ n + 1 1
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1 solution
Even though it is false, we can still split the sum as the entire sum is convergent and the differences between the two sums are finite. For eg : H n = k ≥ 1 ∑ ( k 1 − n + k 1 ) . This is absolutely valid as we are interested in the entire summation rather than individual sums.
There are ample of examples where the difference of two infinity like functions is finite and they are valid. This representation should be valid as the entire sum is convergent.
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Actually, no we cannot.
Have you seen our "Brilliant is down for maintainence" page? It gives an example that you easily extend to "show" that 0 = 1 .
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I can't find that page :( . Yet we have purely logical ways to show 0 = 1 but our logic doesn't allow us to do that. Similarly my argument is just to express the fact that two divergent series can often produce a convergent one. Maybe the representation would be unlikely but it works.
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@Aditya Narayan Sharma – Here's a simple scenario where your logic doesn't apply.
More generally, we can always do something like this when we have two divergent series. This is the underlying reason why we care about absolutely convergent summations.
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@Calvin Lin – Well that's a nice exapmle to improve my thoughts about this. :) . You are right
Even though it is true that n ( n + 1 ) 1 = n 1 − n + 1 1 , we cannot just split up the infinite summation.
Since the sum of reciprocals is The RHS is of the form ∞ − ∞ which is undefined.
The LHS is bounded above by ∑ n 2 1 = 6 π 2 , hence is finite.
Thus, these 2 expresions are not equal to each other.