Theorem 1.1 (Claim) The sum of all natural numbers is

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Theorem 1.1 (Claim) The sum of all natural numbers is − 12
.
We begin with two lemmata.
Lemma 1.2
∞
X
(−1)k =
k=0
1
2
Proof Consider the series
A :=
∞
X
(−1)k = 1 − 1 + 1 − 1 + . . .
k=0
Note that
1−A=1−
∞
X
(−1)k
k=0
= 1 − (1 − 1 + 1 − 1 + . . . )
= 1 − 1 + 1 − 1 + ...
∞
X
(−1)k
=
k=0
=A
Whence
2A = 1
It follows that
A=
1
2
(1)
Lemma 1.3
∞
X
(−1)k+1 k =
k=1
1
4
Proof Let
B :=
∞
X
(−1)k+1 k
k=1
We arrange two copies of this series together spatially like so:
1−2 + 3 − 4 + 5 − 6 + . . .
1 − 2 + 3 − 4 + 5 − ...
And adding term-by-term column-wise, we get
1 + (−2 + 1) + (3 − 2) + (−4 + 3) + . . . = 1 − 1 + 1 − 1 + . . .
=A
From equation (1), we have
2B = A
and therefore
2B =
1
2
B=
1
4
It follows that
(2)
Finally, we are ready for the main result.
Proof of 1.1 Let
S :=
∞
X
k
k=1
Note that
S−B =
∞
X
k=1
k−
∞
X
(−1)k+1 k
k=1
= (1 + 2 + 3 + 4 + 5 + 6 + . . . ) − (1 − 2 + 3 − 4 + 5 − 6 + . . . )
= 0 + 4 + 0 + 8 + 0 + 12 + . . .
= 4 (1 + 2 + 3 + . . . )
= 4S
It follows that
3S = −B
And from equation (2) we have
3S = −
1
4
Thus,
S=−
1
12
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