Infinite geometric series

Consider the series

∑ⁿ_k=1 (2⋅½^k-1) =2+1+½+¼+⅛+⋯
Consider also finding the partial sums for 10, 20 and 100 terms. The sums we are looking for are the partial sums of a geometric series. So,

As the number of terms increases, the partial sum appears to be approaching the number 4. This is no coincidence. In the language of limits,

This type of problem allows us to extend the usual concept of a ‘sum’ of a finite number of terms to make sense of sums in which an infinite number of terms is involved. Such series are called infinite series.

Limits
You may have noticed that in some geometric sequences, the later the term in the sequence, the closer the value is to 0. Another way to describe this is that as n increases, a_n approaches 0. The Value that the terms of a sequence approach, in this case 0, is called the limit of the sequence. Other types of infinite sequences may also have limits. If the terms of a sequence do not approach a unique value, we say that the limit of the sequence does not exist.

Example 1. Write the series

¼+⅛+1/16+1/32+⋯ for n terms
in sigma notation. Notice that the instructions specify that the lower limit of summation should be 1 and the variable k. So that would be a good thing to adhere to.
Solution:
Geometric, with a ratio of ½. Same deal as before; I need a formula for a_k.
a_n=a₁⋅r^n-1=¼⋅½^k-1
Now, this one actually will simplify if I’d like it to, since ¼ is a power of ½. Check this out:
¼⋅½^k-1=½²⋅½^k-1=½^2+k-1=½^k+1.

So there are two different nice ways to write this.

∑ⁿ_k=1 (¼⋅½^k-1) =∑ⁿ_k=1 ½^k+1

One thing to be made clear about infinite series is that they are not true sums! The associative property of addition of real numbers allows us to extend the definition of the sum of two numbers, such as a+b, to three or four or n numbers, but not to an infinite number of numbers. For example, you can add any specific number of 5s together and get a real number, but if you add an infinite number of 5s together, you cannot get a real number! The remarkable thing about infinite series is that, in some cases, such as the example above, the sequence of partial sums (which are true sums) approach a finite limit L. The limit in our example is 4. This we write as

lim_n→∞ ⁡∑ⁿ_k=1 a_k =lim_n→∞ ⁡(a₁+a₂+⋯+a_n)=L.
We say that the series converges to L, and it is convenient to define L as the sum of the infinite series. We use the notation
∑^∞_k=1 a_k =lim_n→∞ ⁡∑ⁿ_k=1 a_k =L.
We can, therefore, write the limit above as
∑^∞_k=1 (2⋅½^k-1) =lim_n→∞ ⁡∑ⁿ_k=1 (2⋅½^k-1) =4.
If the series does not have a limit, it diverges and does not have a sum. We are now ready to develop a general rule for infinite geometric series. As you know, the sum of the geometric series is given by

We will call this the sum of the infinite geometric series. In all other cases the series diverges. The proof is left as an exercise.

, as already shown.

Sum of an infinite geometric series
The sum, S_∞, of an infinite geometric series with first term a₁, such that the common ratio r satisfies the condition |r|<1 is given by:

Looking Ahead to Calculus: Infinite Series

As indicated above, each sequence {a_n} is associated with a sequence of partial sums {S_n}, where S_n=a₁+a₂+⋯+a_n. What happens to S_n as n gets larger and larger, that is, as we add more and more terms? We are considering an “infinite sum” written as a₁+a₂+a₃+⋯, or in summation notation,

∑^∞_n=1 a_n
This is called an infinite series.

Since we cannot add an infinite set of numbers, we need instead the notion of a limit. In one sense, calculus is the study of limits. It is beyond the scope of this book to deal with infinite series in general, but for a geometric sequence {a_n}, we can at least get an intuitive feeling for what happens to S_n as n becomes large. For example, where

it is reasonable to assume that 1/2ⁿ gets close to 0 as n becomes large. In calculus notation lim_n→∞ ⁡1/2ⁿ=0, from which lim_n→∞ ⁡S_n=⅔.

We say that the infinite series,

converges to ⅔, and we write

In general, we associate each geometric sequence {a⋅r^n-1} with an infinite geometric series

∑^∞_n=1 a⋅r^n-1=a+a⋅r+a⋅r²+⋯+a⋅r^n-1+⋯.
The only meaning we give to this infinite sum is the limit of the sequence of partial sums,

which depends on lim_n→∞ ⁡rⁿ. Looking at different values of r, we conclude that if r is any number between -1 and 1, then lim_n→∞ ⁡rⁿ=0, from which

Infinite geometric series
Associated with every geometric sequence { a⋅r^n-1} is an infinite geometric series

∑^∞_n=1 a⋅r^n-1=a+a⋅r+a⋅r²+⋯+a⋅r^n-1+⋯.
If -1<r<1, then the series converges to a/(1-r), and we write
If |r|≥1, then the infinite series does not have a sum, and it diverges.

Example 2. Find the sum

Solution:
We substitute k=1 into the formula 13/100^k and add successive terms until we reach k=4.

Equation-A. Sum of Geometric Sequence:
• The sum S of the first n terms of a geometric sequence a_k=a⋅r^k-1 for k≥1 is

We close this section with a peek into Calculus by considering infinite sums, called series. Consider the number 0.9. We can write this number as

0.9=0.9999…=0.9+0.09+0.009+0.0009+…
From Example 2, we know we can write the sum of the first n of these terms as

Using Equation-A, we have

It stands to reason that 0.9 is the same value of

Our knowledge of any exponential expressions tells us that

We have just argued that 0.9=1, which may cause some distress for some readers.⁷ Any non-terminating decimal can be thought of as an infinite sum whose denominators are the powers of 10, so the phenomenon of adding up infinitely many terms and arriving at a finite number is not as foreign of a concept as it may appear. We end this section with a theorem concerning geometric series.
💎 Let you read a next post, how to express a recurring decimal as a fraction ?
Theorem-A. Geometric Series: Given the sequence a_k=a⋅r^k-1 for k≥1, where |r|<1,
If |r|≥1, the sum a+a⋅r+a⋅r²+⋯ is not defined.

The justification of the result in Theorem-A comes from taking the formula in Equation-A for the sum of the first n terms of a geometric sequence and examining the formula as n→∞,. Assuming |r|<1 means -1<r<1, so rⁿ→0 as n→∞. Hence as n→∞,

As to what goes wrong when |r|≥1, we leave that to Calculus as well, but will explore some cases in the exercises.
_____
⁷ To make this more palatable, it is usually accepted that 0.3=⅓ so that 0.9=3(0.3)=3(⅓)=1. Feel better?
💎 A simpler thought than this page is the next page of Sigma Notation Examples about Infinite Geometric Series.