
sums will set you freethe sum of a geometric sequence:

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Our inflation paper mentions something called “the lender’s multiplier”, which is dependent on the “reserve/deposit ratio”. Here we will explain in more detail what those terms mean and how it works in practice, including the sums. Money circulates from hand to hand. A ‘single’ note or coin is used over and over again. The government pays for real goods, usually through a government bank account. That money will end up at a provider of services, for example at a supplier of motor cars, or as the wages of a government employee. The money that each service provider receives will, in due course, fetch up in another bank account. The bank will then pay out other sums of money at the account holder’s request, sums of money which will, in their turn, end up in other bank accounts. From the point of view of the bank, the situation is similar whether the government deposits the pound, or whether the pound arrives when jo sixpack puts his wages into the bank. A minor difference is that the bank tends to pay more interest to the government than it does to poor old jo. The bank can also borrow money from the government (the Bank of England) at interest. This can include money the government does not have! In other words, the government just prints it. For more details of this clever trick, see our inflation document. Meanwhile, during all these deposits and withdrawals, at any particular time, any bank has on deposit large amounts of inactive money, which the bank is at present not being ordered to pay out. The bank can lend this inactive money to eager borrowers, subject to it retaining a legally prescribed percentage, called the “reserve/deposit ratio”. The reserve/deposit ratio is usually about 10%, meaning the bank can only lend out 90% of its inactive money: 10% must be kept still doing nothing. For illustration, assume the bank has one pound of inactive money, which it can lend. When it lends this money out, by similar routes already described, that 90% (90p) will find its way into another bank (or even the same bank), and the process will repeat itself. Remember the bank still has this one pound on its records as being owned by one of the bank’s depositors, even though a borrower is currently using 90% of it. Now let’s look at fractional banking in terms of arithmetic. To make the simplest possible example, we will work with a reserve/deposit ratio of 10% and we will look at what happens when one bank has one pound, and lends to a second bank, which bank lends to a third, which bank lends to a fourth, and so. Of course in the real world, any two of these banks could be the same bank. We will ignore everything that happens to the money between the borrower borrowing it from the bank and it being put back into a bank by the same borrower, or by someone the borrower paid, or by someone paid by someone the borrower paid, or... Instead, we will work as if the money goes directly from the lending bank to the next depositing bank. In other words, as if the lending bank has lent the money directly to the second bank. The first bank starts with the full amount: one pound. They must keep 10% as a reserve and can lend the rest to the next bank/customer: the rest after reserving 10% is 90% or 0.9, 90p in this case. The next bank takes the 0.9 and must keep 10% of that and can lend the rest: 90% of 0.9 or 0.9*0.9, which is 81p. The next bank takes the 0.9*0.9 and must keep 10% of that and can lend the rest: 90% of 0.9*0.9 or 0.9*0.9*0.9, which is 72.9p. This continues down the sequence of banks, so that the n^{th} bank in the chain receives 0.9 multiplied by itself n1 times (the first bank still has the full pound  100%  on its record as owned by its original depositor, we start 90%ing from first bank where the money is deposited: the second bank in the sequence). This is an example of a geometric [1] sequence. A general geometric sequence can be written as a, ax, ax^{2}, ax^{3},.., ax^{n1} ...where a is the amount you start with, and x is what you multiply by each time. In our example, a is 1 and x is 0.9. We want to know what we get when we add everything in the sequence together: the sum. In other words, we want to know the sum of: 1 + 0.9 + 0.9^{2} + 0.9^{3} + .. + 0.9^{n1} ... Or for the general sequence: a + ax + ax^{2} + ax^{3} + .. + ax^{n1} ... To start with we will only work out the sum for the first n terms, then we will show how to extend that an ‘indefinite’ sequence. Working out the sum involves a trick: we multiple the sum by x (the percentage being lent) and subtract the sum from the result. Here's our sum, we’ll call it S_{n} (the sum of the first n terms): S_{n} = a + ax + ax^{2} + ax^{3} + .. + ax^{n2} + ax^{n1} [1] Now multiply both sides by x, the percentage we are lending out each time: xS_{n} = ax + ax^{2} + ax^{3} + ax^{4} + .. + ax^{n1} + ax^{n} [2] Now subtract [1] from [2]: xS_{n}  S_{n} = (a + ax + ax^{2} + ax^{3} + .. + ax^{n2} + ax^{n1})  (ax + ax^{2} + ax^{3} + ax^{4} + .. + ax^{n1} + ax^{n}) Now you should be able to see that the only thing in xS_{n} that is not in S_{n} is ax^{n}, and the only thing in S_{n} that is not in xS_{n} is a, so we can cancel everything else (the parts in red). If this is not clear to you, write out a few terms of both S_{n} and xS_{n} on some paper and try seeing what you can cancel. Cancelling those terms gives: xS_{n}  S_{n} = ax^{n}  a Factorise both sides: S_{n} * (x  1) = a * (x^{n}  1) Now divide both sides by x1 to get S_{n}:
To make the formula more convenient, we can change the sign on the top and bottom. We can do this because 1 divided by 1 is the same as 1 divided by 1! This gives: S_{n} = a (1  x^{n}) / (1  x) And there we have a general formula for the sum of geometric series with n terms. Let’s try it with our example: we had a=1, x=0.9. We’ll try working out the sum of the first 5 terms: S_{5} = 1 * (1  (0.9)^{5}) / 1  0.9 = (10.59049) / 0.1 = 4.0951 And checking using the way we know works: S_{5} = 1 + 0.9 + 0.9*0.9 + 0.9*0.9*0.9 + 0.9*0.9*0.9*0.9*0.9 + 0.9*0.9*0.9*0.9*0.9 = 1 + 0.9 + 0.81 + 0.729 + 0.6561 = 4.0951 It works, and we now know that with five banks in the chain, the original one pound owned by the first bank has turned into a bit over four pounds. Now what we really want to know is how much that one pound will become if it is lent until the 90%s disappear in a puff of smoke, or the sum to ‘infinity’ (in fact you only have to have a sequence of 45 banks before the last one gets less than one penny from the first bank’s pound). Our sum to n terms is: S_{n} = a (1  x^{n}) / (1  x) In our calculations, x is less than 1: we are taking less than 100% ever time. This means that the more times you multiply x by itself (x^{n} means multiplying x by itself n1 times), the smaller it gets: 0.9*0.9 is less than 0.9, 0.9*0.9*0.9 is less than 0.9*0.9, and so on. Eventually x^{n} will get very small indeed. In other words, the bigger you make n, the smaller x^{n} becomes and if n becomes big enough, x^{n} becomes so small that we can ignore it: it is close enough to nothing that we don’t care. A less careful, or ‘better’ indoctrinated, mathematician would say “x^{n} tends to 0 as n tends to infinity” and make the sign of the cross. In practical terms, this means we can get rid of that x^{n} part of our formula if we are talking about very big n’s  as I said above, with our example, it gets less than a penny in the pound with n as 45 [2]. Removing x^{n} from the formula gives: S_{big_n} = a / (1  x) Or for our example: S_{big_n} = 1 / (1  0.9) = 1 / 0.1 = 10 Thus every pound deposited in the bank becomes ten pounds when we take into account the whole lending chain. That calculation is with a deposit ratio of 10%. With a deposit ratio of 8%, one pound would become twelve and half pounds. With a deposit ratio of 12%, one pound would become eight and a third pounds. So you see that what started off as one pound (still notionally sitting in the original bank) has now become eight, ten or twelve times as much, depending on the reserve ratio set by the government. This is how the original one pound that the government prints ends up causing about ten times as much increase in the actual money stock. This process is known as fractional banking, discovered long ago by fly bankers who realised that they could lend out several times as much gold as they had in their vaults, in the knowledge that not all their customers would want/demand their money at the same time. Note that this goes completely awry in a situation where the public loses faith in the honesty or reliability of a bank, and so attempt to pull out their money as quickly as possible, before the bank runs out of cash. This is known as a “run on the bank”. In such a situation, the embarrassed banker will attempt to call in the loans that he has made, in order to pay off his customers. This is liable to embarrass the next banker, who will likewise start to try to call in his loans, and so on. This is the great fear of many an indoctrinated economist and moonbat, and is known as a systemic banking crisis, crash or panic. This is part of the cause of the 1929 crash, famed in legend and even approximate reality. With modern fiat money, the central bank now just keeps printing money, thus feeding liquidity into the system. In due course it is hoped that the panic will subside, the customers will starting putting their paper back into the banks, and the central bank will start mopping up the excess liquidity. In the crash of 1929, part of the method to stop the problem was simply to close the doors to the bank for some days. In 1929, there was an additional problem: the money was backed by gold and the customers were in a position to claim back their deposits in gold. It was found much more difficult to print gold than to print paper!
Endnotes1. It is the fact that the term is multiplied each time that makes the sequence geometric. In arithmetic sequences, you add something every time, rather than multiplying by something each time. 2. The sum of the first 45 terms comes to 9.9127203643 pounds, very near the final total of 10 pounds: shortchanged by almost 9p. 

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