Why Does Sum of Infinity Equal Infinity?

[SpaceDomain]

I know that

$$\sum_{n=-\infty}^\infty{1} = \infty$$

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

$$\sum_{n=-\infty}^\infty{1}$$

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

$$\sum_{n=-\infty}^\infty{1} = \infty$$

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.