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I have the following series:

$\sum_{n=1}^{\infty} \left(1\over2^{n-1}\right) $

I have to calculate its sum. I don't know how to do so. I'd like to get helped. thanks in advance.

Martin Sleziak
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Billie
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1 Answers1

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Note:

$$\sum_{n=1}^{\infty} \left(1\over2^{n-1}\right) = \sum_{n=0}^{\infty} \left(1\over2^{n}\right) = \sum_{n = 0}^\infty \left(\frac 12\right)^n$$

And so we have a geometric series:

$$ a + ar + ar^2 + ar^3 + ar^4 + \cdots = \sum_{n=0}^\infty \color{blue}{\bf a}\color{red}{\bf r}^n = \frac{\color{blue}{\bf a}}{1-\color{red}{\bf r}} \iff |r| < 1$$

In your case, we have that $a = \color{blue}{\bf 1}$, giving us a sum $\dfrac{1}{1-\color{red}{\bf r}}$, with $\color{red}{\bf r = \dfrac 12} \lt 1$. $$ \sum_{n = 0}^\infty \color{blue}{\bf 1}\cdot \left(\color{red}{\bf \frac 12}\right)^n = \dfrac{1}{1 - \left(\frac 12 \right)} = 2.$$

amWhy
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  • Can you please give me more hint? – Billie May 06 '13 at 18:23
  • I didn't understand the explanation with these r .. can you please explain me that? – Billie May 06 '13 at 18:30
  • Try writing out the first handful of terms: $\left(\frac 12\right)^0 + \left(\frac 12\right)^1 + \left(\frac 12 \right)^2 + \left(\frac 12\right)^3 + \cdots $. What is the common term, call it $r$, that we are summing powers of? – amWhy May 06 '13 at 18:36
  • Sorry but I still don't get it .. – Billie May 06 '13 at 18:37
  • I'm afraid, @user1798362, that if you don't understand this answer and comments you may be needing to go to the basics. It is a loooooong shot to expect someone in this site to explain you something very basic but rather lengthy like geometric progressions/series – DonAntonio May 06 '13 at 18:42
  • @amWhy: Very nice - even color coded! +1 – Amzoti May 07 '13 at 00:40