# Session 6

#### Geometric Series

The following (finite) geometric series pops up in different situations

• S(n) = a + a*a + a*a*a + … + a^n,

where a^n = a*a*a…*a (n times) and it is of interest to find out the value of the sum S(n) as n increases.  We see that the the quotient between consecutive terms is the constant factor a. If a = 1 then S(n) = n and so tends to infinity as n tends to infinity. The same for a>1 with quicker divergence to infinity.

What about 0 < a < 1?  In this case a^n gets smaller and smaller with increasing n and so the series may converge to a finite value.

Consider the special case a = 1/2 with

• S(n) = 1/2 + 1/4 + 1/8 +1/16 +…+ 1/2^n.

See that S(1) = 1/2, S(2) = 1/2+1/4 = 3/4, S(3) = 1/2 + 1/4 + 1/8 = (4 + 2 + 1)/8 = 7/8, S(4) = 1/2 + 1/4 + 1/8 + 1/16 = (8 + 4 + 2 + 1)/16 =15/16, and generally

• S(n) = (2^(n-1) + 2^(n-2) +… + 1)/2^n = (2^n – 1)/2^n= 1 – 1/2^n,

and thus for n very large S(n) = 1 or S(∞) = 1.

Generalise to show that for 0 < a < 1 :

• S(∞) = a + a^2 + a^3 + ….= a/(1-a).

Hint: Use that S(n) =a*(1 + S(n-1)) and S(n) – S(n-1) = a^n to see that

• S(n-1) = (a – a^n)/(1-a)  with a^n tending to zero with increasing n.