Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(|r|<1,\) then \(\sum_{n=1}^{\infty} a r^{n}=\frac{a}{(1-r)} .\)

Short answer

The statement is false because the summation should start from n = 0, not n = 1. This can be demonstrated using a counterexample, such as when \(a = 3\) and \(r = 1/2\), the sum returns an incorrect value if started from n = 1 compared to the correct value when started from n = 0.

Step-by-step solution

Verify the result from Geometric Series Summation

The sum of an infinite geometric series \(\sum_{n=0}^{\infty} a r^{n}\) is given by a / (1 - r), provided that the absolute value of 'r' is less than 1. This equation gives a finite sum when |r| < 1.

Correct the Provided Statement

The given statement is almost correct, but there is a subtle issue related to the starting index of the sum. The correct summation starts from n = 0 not n = 1. This discrepancy makes the statement false.

Provide a Counterexample

For an example, consider \(a = 3\) and \(r = 1/2\). According to the statement, the sum would start from n = 1, hence be \(3/2 + 3/4 + 3/8 + ...\). This isn't equal to \(3 / (1 - (1/2)) = 6\), while if we start from n = 0 giving us \(3 + 3/2 + 3/4 + ...\) which equals 6. Therefore, the statement is false.