The Sum of Geometric Series from Probability Theory

The Sum of Geometric Series from Probability Theory

[Total: 3    Average: 4/5]

Here I present a simple (but to the best of my knowledge, new) derivation of the formula for the sum of the infinite geometric series. The derivation is based on the use of basic probability theory.

Suppose that you play a game (e.g. lottery or roulette) for which the probability of winning is ##p\neq 0##. And suppose that you decide to repeat the game until you win, after which you stop playing. The probability that you will win in the 1st try is ##p_1=p##. The probability that you will win in the 2nd try is the probability that you will first loose once and then win, so the probability  that you will win in the 2nd try is ##p_2=(1-p)p##. The probability that you will win in the 3rd try is the probability that you will first loose twice and then win, so ##p_3=(1-p)^2p##. Hence the probability that sooner or later you will eventually win is

$$P=p_1+p_2+p_3+\cdots = p + (1-p)p+(1-p)^2p + \cdots$$

which can be written as

$$P=p\sum_{k=0}^{\infty}(1-p)^k = (1-q)\sum_{k=0}^{\infty}q^k $$

where ##q\equiv 1-p##. But it is certain that sooner or later you will win, that is, from the theory of probability it is clear that ##P=1##. Hence

$$\sum_{k=0}^{\infty}q^k = \frac{1}{1-q}$$

This result is nothing but the formula for the sum of the geometric series, which I derived here from the theory of probability.

Note that the formula is not valid for ##q>1##, which has an interpretation in probability theory. The quantity ##q=1-p## is the probability of not winning the game, and it does not make sense to have a probability ##q## larger than 1.

 

 

12 replies

Leave a Reply

Want to join the discussion?
Feel free to contribute!

Leave a Reply