Σ2^{-n} = Σn·2^{-n}

Admitedly, this is a strange identity to contemplate: every term on the right is greater than the corresponding term on the left. How then the identity is possible?

The identity becomes more plausible with the remark that the summation on both sides starts with \(n = 0\):

(1) | \(\displaystyle\sum_{n=0}2^{-n}=\sum_{n=0}n2^{-n}.\) |

which effectively kills the first term on the right. This becomes transparent if we write the series explicitly:

(1') | \(1 + 1/2 + 1/4 + 1/8 + ... = 0\cdot 1 + 1\cdot 1/2 + 2\cdot 1/4 + 3\cdot 1/8 + \ldots\) |

Is it now less surprising?

We shall assume that the two series converge. (This is shown by the means taught in the beginning Calculus course. We skip this here. The point is however similar to that made in the discussion of the identity \(0.999 ... = 1\). In the presence of convergence simple algebraic manipulations of a series lead to valid results that may not be true for the divergent series.)

Thus denote

\(S = 1 + 1/2 + 1/4 + 1/8 + \ldots\) and

\(T = 1/2 + 2/4 + 3/8 + \ldots\)

It's actually more convenient to work with the summation symbol \(\sum\):

\( \begin{align} T &= 1/2 + 2/4 + 3/8 + \ldots \\ &= \sum_{n=0}n\cdot 2^{-n} \\ &= \sum_{n=1}n\cdot 2^{-n} \\ &= \sum_{n=1}(n-1)\cdot 2^{-n} + \sum_{n=1}2^{-n} \\ &= \frac{1}{2}\sum_{n=1}(n - 1)\cdot 2^{n-1} + \frac{1}{2}\sum_{n=1}2^{n-1} \\ &= \frac{1}{2}\sum_{n=0}n\cdot 2^{n} + \frac{1}{2}\sum_{n=0}2^{n} \\ &= \frac{1}{2}S + \frac{1}{2}T. \end{align} \)

Which implies

\(S = S/2 + T/2\), or \(S/2 = T/2\).

proving the promised \(S = T\). As \(S\) is the sum of a geometric series with the factor \(q = 1/2\),

\(S = \frac{1}{1 - q} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2.\)

Therefore, \(2 = S = T.\)

The fact just proved has almost practical application [Falk, p. 47]:

A story about a (male) monarch tells that he wished to change the sex ratio distribution in favor of women, so that each man would be able to have a big harem. Unlike ruthless rulers of the past, he chose 'humane' methods to that end. He ruled that every woman could continue to bear babies as long as they were females. However, if a woman gave birth to a boy, she had to stop having babies.

If that regulation was followed, then only certain family types would exist in that population. These are (according to birth order): \(M, FM, FFM, FFFM, \ldots\) Would such a policy, if carried out long enough, fulfill the ruler's expectations? Could it change the sex ratio in favor of males? Or would it fail to change the approximately equal proportions of males and females after all?

Assuming boys and girls pop up with equal probabilites, set \(p=\frac{1}{2}\). The expected number of boys a woman may have is obviously \(1\). This is also the sum of the series

\(1\cdot \frac{1}{2} + 1\cdot (\frac{1}{2})^{2} + 1\cdot (\frac{1}{2})^{3} + \ldots = 1.\)

The expected number of girls a woman may have is the sum of the following series:

\(1\cdot (\frac{1}{2})^{2} + 2\cdot (\frac{1}{2})^{3} + 3\cdot (\frac{1}{2})^{4} + \ldots = 1.\)

For, indeed, as we just saw,

\(\displaystyle\sum_{n=0}2^{-n} = \sum_{n=0}n2^{-n},\)

which shows that, on average, a woman will have one boy and one girl, so that the monarch's regulation will not affect the approximately equal distribution of males and females.

(Note that practically the same derivation applies to finding the expected number of (Bernoulli) trials to first success.)

### References

- R. Falk,
*Understanding Probability and Statistics*, A K Peters, 1993

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