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The sum of all natural numbers is

−

1

12

.

We begin with two lemmata.

Consider the series

A :=

∞

X

(

−

1) k

= 1

−

1 + 1

−

1 + . . .

k =0

Note that

∞

X

(

−

1) k

=

1

2 k =0

1

−

A = 1

−

∞

X

(

−

1) k k =0

= 1

−

(1

−

1 + 1

−

1 + . . .

)

= 1 − 1 + 1 − 1 + . . .

=

∞

X

( − 1) k k =0

= A

Whence

2 A = 1

It follows that

A =

1

2

∞

X

(

−

1) k +1 k =

1

4 k =1

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

−

. . .

(1)

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

2 B = A and therefore

2 B =

1

2

It follows that

B =

1

4

(2)

Finally, we are ready for the main result.

Let

S :=

∞

X k k =1

Note that

S

−

B =

∞

X k

−

∞

X

(

−

1) k +1 k k =1 k =1

= (1 + 2 + 3 + 4 + 5 + 6 + . . .

)

−

(1

−

2 + 3

−

4 + 5

−

6 + . . .

)

= 0 + 4 + 0 + 8 + 0 + 12 + . . .

= 4 (1 + 2 + 3 + . . .

)

= 4 S

It follows that

3 S =

−

B

And from equation (2) we have

3 S = −

1

4

Thus,

S =

1

−

12