Question

A fair coin is tossed repeatedly. The probability that the first head occurs on the \(n\) th toss is given by \(P(n)=\left(\frac{1}{2}\right)^{n},\) where \(n \geq 1\) (a) Show that \(\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n}=1\). (b) The expected number of tosses required until the first head occurs in the experiment is given by \(\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n}\) Is this series geometric? (c) Use a computer algebra system to find the sum in part (b).

Short answer

The sum of the series in part (a) is 1. The series in part (b) is not geometric. The sum of the series in part (b) can be found using a computer algebra system.

Step-by-step solution

Finding whether series in (a) is geometric

Observe that the series \(\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n}\) can be rewritten as \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...\). This suggests the ratio \(r\) between consecutive terms is \(1/2\) as each term is half the previous term.

Computing sum in (a)

Given that series is geometric with first term \(a = 1/2\) and \(r = 1/2\), the sum \(s\) of this infinite series is given by \(s = a / (1 - r)\). Substituting values, \(s = (1/2) / (1 - 1/2) = 1\). Hence, \(\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n} = 1\)

Finding nature of series in (b)

Look at the series: \(\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n}\), Each term could be formed by multiplying \(n\) with \(\left(\frac{1}{2}\right)^n.\) Although each term is halved before proceeding to the next, the \(n\) outside the parentheses breaks the ratio characteristic of a geometric series. So, this is not a geometric series.

Calculating sum in (c)

Using a computer algebra system, input the series \(\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n}\) to compute its sum. Depending upon the system, the method may vary, but typically involves inputting the series using the system’s syntax for infinity and sigma notation.