What is the sum to infinity of geometric progressions?

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In a geometric progression, the sum to infinity can be calculated using the formula a/(1-r), where 'a' is the first term and 'r' is the common ratio. For the series 4, 2, 1 with a common ratio of 1/2, the sum to infinity is exactly 8. As the number of terms approaches infinity, the finite sums get closer to this limit, confirming that the sum is indeed 8. If the sum were less than 8, it would contradict the properties of the real number system. Thus, the sum to infinity is defined as 8, with no ambiguity in its value.
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Hi!
In a geometric progression you can find the sum to infinety is some series, for example 4,2,1,... where the common factor is 1/2. The sum to infinety will then be, 8, it says in my book, but I can only think of it as very, very close to 8, not eight exactly. How is it? Is the sum to infinety 8 or just very close to eight?
/Andreas
 
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Sum of a GP = a(1-r^n)/(1-r) where a is the first number in the sequence, r is the common ratio and n is the term number.

When -1<r<1 and n approaches infinity r^n approaches 0. Therefore Sum to infinity = a/(1-r)

Therefore Sum to infinity = 4/(1-0.5) = 8
 
The sum *is* eight. It is the limit of the finite subsums. If it weren't eight but were less than 8, then yo'ud have a problem since one of the (increasing) finite subsums would be greater than your preferred infinite sum. It is a property of the real number system that the sum is 8. It is, by definition, 8 there is no contention about that, if you think it is something different then you don't understand what the words mean.
 
If you were to stop somewhere short of "infinity", say summing up to n= 10000000, then the answer, one of the "subsums" that matt grime referred to (I would say "partial sum") would be slightly less than 8. Summing all terms, that is, never stopping, will give exactly 8.
 
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