Suppose a light bulb is turned on at time $t=0.$ It switches off at $t=1,$ on again at $t=1+{1 \over 2},$ off at $t=1+ {1 \over 2}+{1 \over 3},$ and so forth. As $t$ goes to $\infty,$ what proportion of the time is the light bulb on?
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4 Answers
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The whole series is the alternating harmonic series, which in the limit as n approaches infinity equals $ln 2$.
As both individual series diverge, the difference goes to a 0 fraction in the limit, and the ratio approaches 1.0. Therefore the light is on half the time.
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Consider the first few terms of the series concerning the period of time the light is turned on and off:
$$\text{ON: }1, \frac 13, \frac 15,...\\ \text{OFF: } \frac 12,\frac 14, \frac 16,...$$
You can clearly see that the times you are looking for are:
$$\displaystyle\sum^{\infty}_{i=1} {\frac 1{2i-1}} $$
and
$$\displaystyle\sum^{\infty}_{i=1} {\frac 1{2i}} $$
Use harmonic series to solve
answered Oct 9, 2013 at 2:45
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$\begingroup$ So....what proportion of the time is the light bulb on? Certainly this diverges. $\endgroup$ Oct 9, 2013 at 2:49
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$\begingroup$ The interesting thing is that the difference of the two series does converge, as the accepted answer points out. $\endgroup$– soakleyOct 9, 2013 at 22:56
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At any point in time, the light has been on longer than it has been off, as $1 > \frac{1}{2}$, $\frac{1}{3} > \frac{1}{4}$, etc. If we start looking at the light at $t = 1$, then at any point in time the light has been off longer than it has been on, as $\frac{1}{2} > \frac{1}{3}$, $\frac{1}{4} > \frac{1}{5}$, etc. So the proportion of the time that the light is on must be $\frac{1}{2}$.
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$\begingroup$ How so? If we denote the proportion of the time that the light is on by $P$, then the first sentence implies $P \geq \frac{1}{2}$ and the second sentence implies $P \leq \frac{1}{2}$. $\endgroup$– ArthurOct 9, 2013 at 4:14
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$\begingroup$ It makes sense to me, but only after I understood Pieter's answer, which isn't that clear on first reading, either. But essentially Arthur is arguing the 1 at the beginning simply doesn't matter, and he is correct. $\endgroup$– soakleyOct 9, 2013 at 22:55
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The switches occur at the following time instants
$$\begin{aligned} t_0 &= 0\\ t_1 &= 1\\ t_2 &= 1 + \frac12\\ t_3 &= 1 + \frac12 + \frac13\\ & \quad \vdots\\ t_n &= \sum_{k=1}^n \frac1k =: h_n\end{aligned}$$
where $h_n$ is the $n$-th harmonic number (where $n \geq 1$). Just after the $n$-th switch, the total amount of time that the light bulb has been on is
$$g_n := \sum_{k=1}^n \left( \frac{1 + (-1)^{k-1}}{2} \right) \frac1k = \frac12 h_n + \frac12 \sum_{k=1}^n \frac{(-1)^{k-1}}{k} $$
where, again, $n \geq 1$. Hence, after the $n$-th switch, the fraction of time that the light bulb has been on is
$$f_n := \frac{g_n}{h_n} = \frac12 \left( 1 + \frac{1}{h_n} \sum_{k=1}^n \frac{(-1)^{k-1}}{k} \right)$$
Since the alternating harmonic series converges,
$$\lim_{n \to \infty} \sum_{k=1}^n \frac{(-1)^{k-1}}{k} = \ln (2)$$
whereas the harmonic series diverges,
$$\displaystyle\lim_{n \to \infty} h_n = \infty$$
we conclude
$$\lim_{n \to \infty} f_n = \frac12 + \frac12 \underbrace{\left( \lim_{n \to \infty} \frac{1}{h_n} \sum_{k=1}^n \frac{(-1)^{k-1}}{k} \right)}_{= \frac{\ln (2)}{\infty} = 0} = \frac12 + 0 = \color{blue}{\frac12}$$