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

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SUMMARY

The sum of the infinite geometric series \(\sum_{n = 0}^{\infty}\frac{1}{2^{n}}\) converges to 2, as established through the properties of geometric series. The series can be expressed as \(\frac{1}{2^{0}} + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \cdots\), which approaches 2 without extensive calculations. This conclusion is supported by the method suggested by Ackbach, demonstrating that while many convergent series are not computable, this particular series is well-behaved and computable.

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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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