So, infinite sums. We covered these. These are an infinitely long list of numbers which you add up. If we want to add up all of the positive whole numbers we would write out:
\[\sum_{x=1}^\infty x\]
If we wanted to add up all the even numbers we would write out:
\[\sum_{x=1}^\infty 2\cdot x\]
Of course, both of these additions is a pointless exercise, because the answers are themselves infinite. There are an infinite amount of whole numbers, and if you add them all up you get infinity. Hell, even if you just added one to itself an infinite amount of times you would also get infinity:
\[\sum_{x=1}^\infty \tfrac{x}{x}\]
(Any number divided by itself is automatically one).
Does this always hold true? Do infinite sums ALWAYS add up to infinity? What about this sum:
\[\sum_{x=0}^\infty \frac{1}{2^x}\]
Looks a bit more complicated doesn't it? First of all notice that I am going to start adding from the zero'th position in the series. So first of all I plug in 0 for \(x\). I get one divided by two to the power of zero. Remember that anything to the power of zero is one. So the first term in this series is one divided by one, or one.
The next term is one divided by two to the power of one. Anything to the power of one is a just one copy of itself. So this is just two. So the term is one divided by two or a half. So far our sum is one plus a half.
The next term is one divided by two squared. Two squared is four, so this is one quarter. The next term is going to be one over two cubed, or an eighth and so on. Basically the series is one plus a whole long list of the inverses of the powers of two. Looks a bit like this:
\[1+\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\tfrac{1}{32}+\tfrac{1}{64}+\tfrac{1}{128}+\tfrac{1}{256}+\tfrac{1}{512}+\ldots\]
Does that add up to infinity as well? Hmm. Maybe not - look at each term, they all get smaller very very quickly. If they get smaller quickly enough, then adding them all up may not reach infinity.
Lets try some mathematical wizardry. Lets create a variable, which we will call \(x\). Actually, no, lets call it \(S\) for series instead. now let us give make the variable equal to the series that we have created from our sum. To keep things simple we'll cut down the number of terms on display:
\[S=1+\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\ldots\]
OK, now we will divide each side by two:
\[\frac{S}{2}=\frac{1+\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\ldots}{2}\]
If we divide a long series of additions by two, that is the same as dividing each individual number in the series by two. Starting with the one at the beginning, that will become a half, and then the half becomes a quarter and the quarter an eighth and so on. Can you see that we are effectively just throwing away the one, and moving every other entry in the series one to the left. Because the series is infinitely long, it is still infinitely long because infinity minus one is still infinity. So once we have done all our divisions by two we get:
\[\frac{S}{2}=\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{1}{16}+\ldots\]
OK. What can we do with that then? Look up two equations to the one where we first defined what \(S\) was going to be. You can see that it is one plus a string of fractions. Now look at the equation above. We have established that one half of \(S\) is a string of fractions. Now consider the string of fractions itself. It looks similar. In fact it looks identical, and it will remain identical no matter how many terms you write down for either one. So far, so good. We can now say that \(S\) is one plus that string of fractions, but we know that the string of fractions is actually \(\tfrac{S}{2}\) so we can actually write:
\[S=1+\frac{S}{2}\]
And if we subtract one half of \(S\) from both sides we get:
\[S-\frac{S}{2}=1\]
\(S\) minus half of \(S\) is obviously just the other half of \(S\), so:
\[\frac{S}{2}=1\]
\[S=1\cdot2\]
\[S=2\]
So we have proved that the whole series up there adds up to two, even though there are an infinite amount of terms in it.
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