So, without further ado, here it is:
As we may intuitively think, the statement about this series seems accurate:
$$1+2+4+8+16+...=\sum_{i=0}^{\infty}2^i = \infty$$
...but is it really \(\infty\)? Let's break it down: we can multiply the same sum by 1 without altering the result, so
$$1+2+4+8+16+... = 1\times(1+2+4+8+16+...)$$
Since \(1 = 2-1\), we can replace it in the equation above like so:
$$1\times(1+2+4+8+16+...) = (2-1)\times(1+2+4+8+16+...)$$
Pretty basic right? Let us now apply the distributive property to get this:
$$(2-1)\times(1+2+4+8+16+...) = 2\times(1+2+4+8+16+...)-1\times(1+2+4+8+16+...)$$
...and if we get rid of the parenthesis we'll get something like this:
$$= 2+4+8+16+32+... -1-2-4-8-16-32-...$$
You may have noticed already that all the numbers in the left, cancel out with all numbers in the right, except for a remaining -1:
$$ \not{2}+4+8+16+32+... -1-\not{2}-4-8-16-32-...$$
$$ \not{2}+\not{4}+8+16+32+... -1-\not{2}-\not{4}-8-16-32-...$$
$$ \not{2}+\not{4}+\not{8}+16+32+... -1-\not{2}-\not{4}-\not{8}-16-32-... = -1$$
Since we come from a series of equalities, we can safely conclude that:
$$\sum_{i=0}^{\infty}2^i =-1$$
Fun, isn't it?
If the intuitive notation for the series \((1+2+4+8+...)\) bothers you, you can replace it by any name you want, lets say \(S\):
$$1+2+4+8+... = \sum_{i=0}^{\infty}2^i = S$$
An alternative way to get the same result is to take the common factor 2 out of the series (starting from the second term of course) like so:
$$S = 1+2+4+8+16+... = 1 + 2(1+2+4+8+...) = 1 + 2S$$
and from here is just a matter of letting the \(S\) on both sides cancel out and you'll get a nice contradiction:
$$S = 1+2S$$
$$2S - S = -1$$
$$S = -1$$
...and that's how it's done.
As we may intuitively think, the statement about this series seems accurate:
$$1+2+4+8+16+...=\sum_{i=0}^{\infty}2^i = \infty$$
...but is it really \(\infty\)? Let's break it down: we can multiply the same sum by 1 without altering the result, so
$$1+2+4+8+16+... = 1\times(1+2+4+8+16+...)$$
Since \(1 = 2-1\), we can replace it in the equation above like so:
$$1\times(1+2+4+8+16+...) = (2-1)\times(1+2+4+8+16+...)$$
Pretty basic right? Let us now apply the distributive property to get this:
$$(2-1)\times(1+2+4+8+16+...) = 2\times(1+2+4+8+16+...)-1\times(1+2+4+8+16+...)$$
...and if we get rid of the parenthesis we'll get something like this:
$$= 2+4+8+16+32+... -1-2-4-8-16-32-...$$
You may have noticed already that all the numbers in the left, cancel out with all numbers in the right, except for a remaining -1:
$$ \not{2}+4+8+16+32+... -1-\not{2}-4-8-16-32-...$$
$$ \not{2}+\not{4}+8+16+32+... -1-\not{2}-\not{4}-8-16-32-...$$
$$ \not{2}+\not{4}+\not{8}+16+32+... -1-\not{2}-\not{4}-\not{8}-16-32-... = -1$$
Since we come from a series of equalities, we can safely conclude that:
$$\sum_{i=0}^{\infty}2^i =-1$$
Fun, isn't it?
If the intuitive notation for the series \((1+2+4+8+...)\) bothers you, you can replace it by any name you want, lets say \(S\):
$$1+2+4+8+... = \sum_{i=0}^{\infty}2^i = S$$
An alternative way to get the same result is to take the common factor 2 out of the series (starting from the second term of course) like so:
$$S = 1+2+4+8+16+... = 1 + 2(1+2+4+8+...) = 1 + 2S$$
and from here is just a matter of letting the \(S\) on both sides cancel out and you'll get a nice contradiction:
$$S = 1+2S$$
$$2S - S = -1$$
$$S = -1$$
...and that's how it's done.
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