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How to estimate summations

daigo

Member
Jun 27, 2012
60
i.e.

[tex]\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...[/tex]

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?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192

CaptainBlack

Well-known member
Jan 26, 2012
890
i.e.

[tex]\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...[/tex]

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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