MTH632 Final Term Topic Lectures 55-63

Topic 055

Geometric Series

Geometric series is one of the most appearing example of series in mathematics and hence it has applications in many areas of science. This is the most basic example. We are choosing this example to understand the concepts of convergence and divergence. So it is very fascinating idea that we are adding infinitely many terms and still if the series is convergent we are going to get some fixed complex number. Now this example is going to give us this idea in much simple way and we will be able to observe this phenomena that how the addition of infinitely many numbers is going to give us fixed complex number and off course in some cases it is also going to be divergent as well. Geometric Series: An infinite series of the form ∑ 𝑛=0 is called geometric series.

Examples of geometric series:

z = 1 ⟹ ∑ (1)𝑛 𝑛=0 z = i ⟹ ∑ 𝑛=0 z = 2 + i ⟹ ∑ 𝑛=0 Now the question is for what values of z geometric series 𝑛=0 converges or diverges? Consider the geometric series:

𝑛=0 = 1 , z , z2 , z3 , …., zn,…. The terms of the series are 1 , z , z2 , z3 , …., zn,…. Since the ratio of any two consecutive terms is z , so z is called ratio or common ratio.

Theorem: If |z| < 1 , the series ∑ 𝑛=0 1−𝑧.

𝑛=0 = 1 + z + z2 + z3 + ….+ zn +…. = 1−𝑧 .

We have to show that:

𝑛=0 = 1 + z + z2 + z3 + ….+ zn +…. = 1−𝑧 .

∞ is sequence of partial sums. Sn = 1 + z + z2 +….+ zn-1 zSn = z + z2 + …. + zn-1 + zn Sn – zSn = 1 - zn

Sn = 1−𝑧

Sn = 1−𝑧 − 1−𝑧 lim 𝑛→∞ 1−𝑧 - lim 𝑛→∞ 1−𝑧

lim 1−𝑧 - 0 = 1−𝑧 Proved!

𝑛=0 If ∑ 𝑛=0

𝑛=0 is divergent.

We have to show that if n ≥ 1 , then ∑ 𝑛=0 is divergent.

𝑛=0 is divergent. We have proved in much simple way but in fact we have proved non-trivial and important result that when we add infinitely many complex numbers then in some cases it is going to some fixed number and we also calculated the fixed number. In convergent case it is going to be 1−𝑧

Following steps of previous proof we can show that

1−𝑧 if |z| < 1

In general sum of geometric series with common ratio z is

1−𝑧 if |z| < 1