Why Does Sum of Infinity Equal Infinity?

[SpaceDomain]

I know that

<br /> \sum_{n=-\infty}^\infty{1} = \infty <br />

But I don't understand why.

It seems to me that since the constant inside the summation is not dependent upon n it can be moved outside the summation. Then there is nothing to sum. It seems to me that

<br /> \sum_{n=-\infty}^\infty{1} <br />

should equal 1.

What am I missing?

 

[mathman]

Your logic is incorrect. The sum is 1+1+1+1+... no matter how you slice it.

You can take the 1 outside the sum, but you still have 1 inside, not 0. 1x0=0, not =1.

 

[SpaceDomain]

I thought the summation just vanished if there was no "argument" inside it.

That makes a lot more sense of why the summation is a discrete analogue of an integral.

Thank you for the help.

 

[SpaceDomain]

Hey mathman, can you go into the other post I posted here:
https://www.physicsforums.com/showthread.php?p=2855526#post2855526

It is basically the same problem, but I still am not understanding the logic. Thanks.

 

[CRGreathouse]

I know that

<br /> \sum_{n=-\infty}^\infty{1} = \infty <br />

But I don't understand why.

It seems to me that since the constant inside the summation is not dependent upon n it can be moved outside the summation. Then there is nothing to sum.

You can change
\sum_{n=1}^N1
to
(N\cdot1)+\sum_{n=1}^N0=N
by 'pulling out the 1'. In your infinite sum, proceeding formally, this would give you
\sum_{n=1}^\infty1
to
(\infty\cdot1)+\sum_{n=1}^\infty0=\infty
which shows (in a non-rigorous way) that the sum diverges.