1. Consider the infinite sum:

1/2 + 1/4 + 1/8 + ... + 1/2Intuitively, it won't go above 1, and will be higher than any number below 1. So, we might take it to be 1.^{n}+ 1/2^{n+1}...

2. Now consider:

1-1+1-1+1-1+...One way to think about this sequence is this:

(1-1)+(1-1)+(1-1)+...This way, it becomes 0+0+0+..., and hence it should equal 0.

3. Another way to break it down is to observe that -1+1=-(1-1) and substitute accordingly:

1-(1-1)-(1-1)-(1-1)...But if we count this way, we get 1-0-0-0-0.... = 1.

4. Now, call the whole sum U.

U= 1-1+1-1+1-1+...The reverse: -U will be obtained by reversing the signs before the summands:

-U= -1+1-1+1-1+1...Add now 1 to each side:

1-U= 1-1+1-1+1-1+1...But this is nothing but 1-U=U! Thus, it should follow that U=1/2.

The question some 18th-century mathematicians asked: which one is it?

## 2 comments:

That particular series troubled me quite a bit until I was taught in my undergraduate calculus course about the concepts of partial sums and their limit. Didn't Euler have to do something with the answer U = 1/2?

Right, so nowadays we just say that there's no such a sum because the sequence doesn't converge.

Leibniz got U=1/2 by expanding this as a power series:

1/(1+x)=1-x+x^2-x^3+x^4-...

Substitute 1 for x, on the right you get the sequence, on the left 1/2.

Euler later used the same method, but he also expanded:

1/(1-x)=1+x+x^2+x^3+...

This of course is problematic, e.g. if you take x to be 2 you get:

-1=1+2+4+8+...

Still, Euler defended sums of divergent series for practical reasons.

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