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### 1 + 2 + 3 + 4 + ... Equals ... -1/12?!

```Date: 09/18/2012 at 14:07:21
From: sankar
Subject: divergent series

1 + 2 + 3 + 4 + ..... = -1/12

How is this possible?

```

```

Date: 09/23/2012 at 21:14:37
From: Doctor Vogler
Subject: Re: divergent series

Hi Sankar,

Thanks for writing to Dr. Math.

This is a common point of confusion. Allow me to explain what's going on
here.

Recall that in order for a sequence to converge, the terms must get
smaller and smaller and go to zero.

But the converse is NOT true: Just because the terms go to zero does NOT
imply that the series converges.

For example, it turns out that this series is very close to (within 1 of)
the natural logarithm of n (that is, ln n):

1 + 1/2 + 1/3 + 1/4 + ... + 1/n

It increases more and more slowly as n gets bigger, but it keeps
increasing, and ln n grows to infinity.

So you are correct that this grows to infinity, or diverges:

1 + 2 + 3 + ...

Now, it seems pretty bizarre to suggest that it could be a negative
fraction, namely -1/12.

But it turns out that this function can be extended in a unique way --
known as analytic continuation -- to all complex numbers except s = 1
(where the function grows to infinity):

zeta(s) = 1/1^s + 1/2^s + 1/3^s + 1/4^s + ...

This converges for s > 1, and for all complex numbers s the real part of
which is bigger than 1. And *this* function has the value -1/12 when

http://mathforum.org/library/drmath/view/52831.html

This is much like summing the series ...

1 + x + x^2 + x^3 + x^4 + ...

... and getting ...

1/(1 - x)

... and then evaluating this at x = 2 to get

1 + 2 + 4 + 8 + 16 + ... = -1.

It's ludicrous, because the series doesn't converge at x = 2. But it
converges (for x < 1) to a function that *does* have a value when x = 2.
That's essentially what happens with analytic continuation. (Indeed, if
you treat x as a complex number, then analytic continuation extends the
geometric series to the fraction I mentioned.)

http://en.wikipedia.org/wiki/Analytic_continuation

This, too:

http://en.wikipedia.org/wiki/Riemann_zeta_function

and show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Analysis
High School Analysis
High School Sequences, Series

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