An apparent contradiction in various explanations of frequency bands and data bandwidth From wikipedia: 
"A key characteristic of bandwidth is that any band of a given width can carry the same amount of information, regardless of where that band is located in the frequency spectrum. For example, a 3 kHz band can carry a telephone conversation whether that band is at baseband (as in a POTS telephone line) or modulated to some higher frequency."
(my emphasis)
From a previous post on this site:
Why is channel capacity a factor of bandwidth instead of frequency?
Question extract: "According to this formula [Shannon-Hartley], a fixed-frequency signal would have the same performance regardless of whether it's at high or low frequency..[]... For example say my bandwidth is 1Hz at a fixed frequency of 1Hz. Compare this with a bandwidth of 1Hz at a frequency of 2.4GHz. It's plainly obvious that I can cram way more bits into 2.4 x 109 cycles/second than I can with just 1/sec."
Extract from one answer: "Certainly you could". 
(note that the answer goes on to discuss signal modulation and energy, at which point I'm lost). 
Question
I'm in way over my head on this topic, and the list of things I don't understand is embarrassingly long, despite having carefully read through the many related posts on this site. 
However, the Wikipedia entry seems intuitively wrong. If we are using electro-magnetic waves to transmit 1s and 0s, then more waves per second must surely translate into a higher rate of 1s and 0s transmitted. So, to my simple mind, a 3 KHz band at 2.4 GHz must be able to carry  more 1s and 0s per second than a 3 KHz band at say 4 KHz.  
Is there any way of explaining this to someone who studied economics, rather than engineering or physics?
 A: 
So, to my simple mind, a 3 KHz band at 2.4 GHz must be able to carry more 1s and 0s per second than a 3 KHz band at say 4 KHz.

Say you have a modulation scheme that provides 1 bit per second per hertz of bandwidth used.
If you use this modulation on a carrier at 2.4 GHz to send 3 kb/s, it will create a signal with a spectrum about 3 kHz wide. If you use it to send 10 kb/s, it will create a signal with a spectrum about 10 kHz wide. etc.
If you use this modulation on a carrier at 4 kHz to send 3 kb/s, it will create a signal with spectrum about 3 kHz wide. If you try to use it to send 10 kb/s, it will fail because you need to make a spectrum 10 kHz wide, but with a carrier at 4 kHz, this means the negative and positive spectra overlap in the frequency band from -1 to +1 kHz. 
So if you have a higher-frequency carrier (or center frequency if your modulation scheme doesn't have a carrier), you can send higher data rates. But doing so creates a signal with wide bandwidth. 
Edit
Here's an example of a simple modulation scheme: To send a one, send the carrier for 1 us. To send a zero, turn the carrier off for 1 us. 
It doesn't matter if the carrier is 10 MHz or 10 GHz, or 350 THz, it takes one microsecond to send one bit, so the bit rate is 1 Mb/s. 
And, whether the carrier is 10 MHz or 10 GHz, the spectrum of the modulated signal will have a peak with about 1 MHz spread. But the center frequency will be equal to the carrier frequency.
If you changed the modulation scheme to turn the carrier on or off for only 0.5 us for each bit, you'd get 2 Mb/s bit rate, and the spread of the spectrum around the carrier would be about 2 MHz instead of 1 MHz. Again, it wouldn't matter what the carrier frequency is.
A: Let's consider amplitude modulation. Consider a carrier at frequency $f_0$. Then the signal will look like $S(t) = A(t)\cos(2\pi f_0 t)$. When we perform amplitude modulation it means we control $A(t)$, the amplitude of the carrier, in such a way that $A(t)$ represents the signal we are trying to send. Let's consider a simple 1s and 0s scheme. Say if $A(t) = 10$ we consider the signal to be a 1 at time $t$ and if $A(t) = 5$ we consider the signal to be 0 at time $t$.
The question is now how fast can we switch the signal, $A(t)$ from 0 to 1. I think you're misconception is that we can switch $A(t)$ from zero to one within the period of the carrier, $T_0 = \frac{1}{f_0}$. Let's explore this for a minute.
Misconception Exploration
Let $A(t) = 0$ represent at 0 and $A(t)=10$ represent a 1. We could then send the signal $S(t) = 10 \cos(2\pi f_0t)$. Then at $t=0$ the signal would be a 1 and at $t = \frac{1}{4}T_0$ the signal would be a 0 and basically the signal would switch from 0 to 1 a couple of times per carrier period. However, if this was our method of sending signals then we couldn't send very interesting signals. Every signal would just be 010101010101.... The point is that what I have just describe here is NOT how signals are sent and does NOT represent how amplitude modulation (AM) works. 
AM bandwidth
Let's return to the previous setup where $A(t) = 10$ is a 1 and $A(t) = 5$ is a 0.
So if the bandwidth, $f_{BW}$ is not equal to the carrier frequency, $f_0$ what is it related to? Let's look into it. Like I said before $A(t)$ is going to carry our signal so we need to look into more detail at $A(t)$. Let's consider a signal $A(t)$ which is a step function. for $t=0$ to $t=10T_0$ we will set $A(t) = 10$ and for $t=10 T_0$ to $t=20 T_0$ we will let $A(t)=5$. What someone receiving this signal would then see is a cosine signal with amplitude of 10 for 10 periods and 5 for the next 10 periods. They would then interpret the signal as being 1 for the first 10 periods and 0 for the second 10 periods. It took $20T$ to send 2 bits then. We can then consider a bitrate $\frac{2 \text{bits}}{20 T} = \frac{1 \text{bit}}{10 T} = 0.1 f_0$. This bit rate is very closely related to the signal bandwidth. It will turn out that they're basically the same thing. I'll get to that in a little bit. But first, let's consider another scenario where we keep the signal high for 5 periods and then switch it low for 5 periods. In this case we were able to send 2 bits in only $10T_0$ so the bit rate is twice as fast, $0.2f_0$. We see that if we can switch the signal faster the bit rate goes up.
Bitrate vs. bandwidth
So far I've explained bitrate and we see that it has to do with how quickly we change $A(t)$, but I haven't related this to bandwidth at all. I think this is at the heart of your question. What is the relationship between bit rate and bandwidth. In fact, even more to the heart of your question is probably the question what is the bandwidth of a signal?
It turns out that we should think of the bandwidth as the inverse of how quickly we change the signal from one state to another state. So, for example, in the first example I gave above we were able to switch the signal from a 1 to a 0 after $10T_0$ so by the definition I just gave the bandwidth of that signal would be $f_{BW}\frac{1}{10T_0} = 0.1 f_0$. Ok, so I've basically just defined the bandwidth to be the same as the bitrate. The question then is why do we call it bandwidth and not bitrate? 
Fourier transform interlude
The answer has to do with Fourier Transforms. If you're not familiar with Fourier transforms here is the quickest intro in the world. When we take the Fourier transform of a signal $S(t)$ we basically break the signal up into different 'components' at different frequencies and define a new function $\tilde{S}(f)$ which tells us how strong our signal is at different frequencies. So for example, if we take the fourier transform of a song with very strong bass (low frequency) then $\tilde{S}(f)$ will be high at low frequencies and lower at high frequencies whereas if we take the Fourier transform of a song with a lot of treble the fourier transform will be high at high $f$ and lower at low $f$.
Fourier transform and signal changing speed
In signal processing there is a very important relationship between frequency and time. Whenever a signal is high frequency that means it is changing very fast meaning the time for the signal to change is very short. Hammer that into head. High frequency means short switching time.
Let's now think about $A(t)$. Remember $A(t)$ carries the information we are interested in. We want to take the Fourier transform of $A(t)$ to get $\tilde{A}(f)$, the frequency spectrum.
Here is the heart of the answer to the heart of your question. It turns out (look up more about Fourier transform) that if your signal changes on a time scale of $T_a$ then the Fourier transform will have components up to $f_a = 1/T_a$. For example, in the first example up above $T_a = 10 T_0$ so $f_a = 0.1 f_0$. This is what we call the bandwidth of a signal. The bandwidth of a signal is the largest frequency in the fourier transform of a signal, $f_{BW} = f_a$. So we apply this to all the examples from before so that $f_{BW} = 0.1 f_0$ as claimed earlier.
Consider the bass song and the treble song, we can see that the bandwidth of the bass song would be lower than the bandwidth of the treble song since the treble song needs to go up to higher frequencies.
Now we know what bandwidth is.
Carrier
But what about the carrier? I've totally stopped talking about the carrier, what happened? Ok, let's go back to that. Let's think JUST about the carrier, $C(t) = \cos(2\pi f_0 t)$. What does the Fourier transform look like? Well in time it only has a signal at 1 frequency so the fourier transform $\tilde{C}(f)$ is basically going to be a spike at $f=f_0$ and have no strength anywhere else. So let's recap our two fourier transforms. $\tilde{C}(f)$ has a spike at $f=f_0$ and $\tilde{A}(f)$ is non-zero from $f=0$ up until $f=f_{bw}$.
We now want to consider the Fourier transform of the actual signal $S(t) = A(t)C(t)$. It turns out* that when we look at the fourier transform of our carrier modulated with our signal it will look exactly like $\tilde{A}(f)$ but with the whole thing shifted up in frequency so that instead of being close to 0 with a width of $f_{bw}$ it will be close to $f_0$ with a width of $f_{bw}$. We say "the signal has been mixed up to the carrier frequency". The key point is that the width of the signal in Fourier space, the "band width" is still $f_{bw}$ even though it is now centered at the carrier frequency $f_0$.
Summary
To summarize the answer to your question:
1) The bitrate is how quickly we can change the signal $A(t)$.
2) How quickly a signal changes in time, $T_a$ is related to the inverse of the highest fourier frequency, $\frac{1}{f_a}$
3) We call this highest frequency $f_a$ the bandwidth, $f_{BW}$ since it represents the width (in frequency) of the Fourier spectrum.
4) Thus the bandwidth is equal to the bitrate.
5) When we modulate a carrier with a signal then the total signal $S(t)$ has the same bandwidth as the information signal, $A(t)$.
With the net result that if you want to change your modulated signal very quickly you need a high bandwidth, noting that this whole discussion does not depend on the carrier frequency, but rather the highest frequency in the Fourier spectrum of $A(t)$.
Caveat
I will note ONE final point. The signal bandwidth, $f_{BW}$ must be less than the carrier frequency $f_0$. So if we have a 10 MHz carrier it is not possible to have a 100 MHz bandwidth, whereas if we have a 1 GHz carrier it is possible to have a 100 MHz bandwidth. This means that for higher frequency signals we technically can cram in more information. HOWEVER, for a high frequency signal and a low frequency signal with the SAME bandwidth, you can cram in the same information rate. So a 1 GHz carrier with 1 MHz bandwidth and 10 MHz carrier with 1 MHz bandwidth would be able to transfer a signal at the same rate.
*Fourier Convolution Theorem
A: While carrier frequency and bandwidth are theoretically separate (as long as bandwidth is less than carrier!), there are practical reasons why high bandwidths and high carrier frequency tend to go together:
1. Antennas, filters and amplifiers in practical radios operate best when the range of frequencies they operate over is small compared to the carrier. Wide band designs do exist (e.g. ultrawideband "UWB" links) but they are noisier, difficult and expensive.
2. As higher frequencies become exploitable with advances in technology, there are many applications to share. A century ago, you might have tried to commandeer the entire AM band for a high speed communications system but it would be unpopular on many levels. (Again, UWB is an exception, but only as a very low power short personal link.)
A: Without getting into complicated math, the general idea is that the carrier is not what transmits information. A pure sine wave at 10 GHz does not transmit information (well, maybe a few bits if you decided before sending it that amplitude and/ or frequency mean something, but that is besides the point here). What transmits information is what you do to the carrier wave. This can be amplitude modifications, frequency modifications, etc...These modifications are called modulations in technical terms. 
So, when you see

It's plainly obvious that I can cram way more bits into 2.4 x 109
  cycles/second than I can with just 1/sec

It just means that if you want to, you can transmit more information with the high frequency carrier than with the low frequency carrier. It doesn't mean that you will do this all the time. You can very well transmit 3kHz of bandwidth on a 2.4GHz carrier.
Let's take an extreme example. You have a light source, say a LED, and you transmit Morse code to your friend 10 m away. The carrier frequency (frequency of the light) is very high,of the order of $10^{14}$Hz. If you do the Morse code modulation manually, by hiding or showing the light with the help of your hand, you may be able to achieve a letter every few seconds. If you use electronics instead at the source and an electronic receptor to turn on and off the LED very rapidly and receive these on/off signals to interpret them, you can transmit many more letters. The number of letters transmitted is your bandwidth. The light is your carrier. 
These things are very difficult to understand deeply without using the appropriate math. My understanding is that economists are not afraid of math, so your best bet would be to understand Fourier series and transforms if you really want to dig into this. They may seem a bit difficult, but they only require prior knowledge of integrals and derivative to get into, which you probably have. Than, it's a question of studying the new concepts, which are not that difficult. 
Also, while this site and Wikipedia are useful, a good structured course/book/tutorial is always useful. You can find a lot for free of free sites/PDFs on the web about these subjects. Unfortunately, I don't have anything specific to recommend as I studied all this 30 years ago and would not know of a good one online. 
