How to estimate summations

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July 4th 2012, 03:22 AM
daigo

How to estimate summations

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?
July 4th 2012, 03:27 AM
Prove It

Re: How to estimate summations

Quote:

Originally Posted by daigo

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?

Yes, this is an infinite geometric series with $\displaystyle \begin{align*} a = 1, r = \frac{1}{2} \end{align*}$ , so the sum is $\displaystyle \begin{align*} \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \end{align*}$ .
July 4th 2012, 07:49 AM
richard1234

Re: How to estimate summations

The sum of a geometric series is given by

$\sum_{i=0}^{\infty} r^i = \frac{1}{1-r}$ , given $|r| < 1$ . In this case, $r = \frac{1}{2}$ .