MHB How Can You Determine the Sum of an Infinite Geometric Series?

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The discussion focuses on determining the sum of an infinite geometric series, specifically the series where each term is half of the previous one. It highlights that while there is no universal method for calculating the sum of all convergent series, certain well-behaved series, like the one presented, can be summed easily. The series in question converges to 2, and the method suggested by Ackbach is mentioned as a way to find this sum without extensive calculations. The conversation emphasizes the computability theory aspect, noting that most series do not have computable sums. Ultimately, understanding the properties of geometric series can simplify the process of finding their sums.
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i.e.

\sum_{n = 0}^{\infty}\frac{1}{2^{n}} = \frac{1}{2^{0}} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}} + \frac{1}{2^{5}} + \frac{1}{2^{6}} + \cdots = ~1.99138889...

Is there a way you can know this solution is 2 without having to perform all of the calculations I did to find which number the sums are approaching? And is there a general method for questions like these to find the solution without having to perform a lot of calculations?
 
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daigo said:
i.e.

\sum_{n = 0}^{\infty}\frac{1}{2^{n}} = \frac{1}{2^{0}} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}} + \frac{1}{2^{5}} + \frac{1}{2^{6}} + \cdots = ~1.99138889...

Is there a way you can know this solution is 2 without having to perform all of the calculations I did to find which number the sums are approaching? And is there a general method for questions like these to find the solution without having to perform a lot of calculations?

There is no general method to determine the sum of a convergent series, it is a result of computability theory that almost all such series are not even computable. This one however is well behaved and its sum can be found using the method sugested by Ackbach

CB
 
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