Why does the summation of a constant from -N to N equal 2N+1?

[SpaceDomain]

I know that:

<br /> \sum_{n=-N}^N{1} = 2N+1 <br />

But I don't understand why.

It seems to me that since the constant inside the summation is not dependent on n it can be moved outside of the summation, leaving nothing to sum.

So I think the summation should actually equal one.

Could someone help me figure out where my logic isn't right?

 

[ehild]

I know that:

<br /> \sum_{n=-N}^N{1} = 2N+1 <br />

But I don't understand why.

It seems to me that since the constant inside the summation is not dependent on n it can be moved outside of the summation, leaving nothing to sum.

If you factor out a number from a sum, like 2x+2y+2z=2(x+y+z), you divide all terms with the common factor. It has no sense to factor out "1" from the summation, as dividing by one leaves everything unchanged.The sum means that you have to add 2N+1 "1" together. It is like counting from -N to N .


ehild

 

[berkeman]

I know that:

<br /> \sum_{n=-N}^N{1} = 2N+1 <br />

But I don't understand why.

It seems to me that since the constant inside the summation is not dependent on n it can be moved outside of the summation, leaving nothing to sum.

So I think the summation should actually equal one.

Could someone help me figure out where my logic isn't right?

Even if you "move" the 1 out in front of the summation, you still leave a 1 inside. You don't leave a 0 inside. Think of the 1 inside as 1 * 1. You bring a constant 1 out in front of the summation, but that still leaves a 1 inside. Make sense?


EDIT -- beaten out by ehild...

 

[Mark44]

I know that:

<br /> \sum_{n=-N}^N{1} = 2N+1 <br />

But I don't understand why.

It seems to me that since the constant inside the summation is not dependent on n it can be moved outside of the summation, leaving nothing to sum.

So I think the summation should actually equal one.

Could someone help me figure out where my logic isn't right?

If the summation were
\sum_{n=1}^5{1}
can you see that it would be 1 + 1 + 1 + 1 + 1 = 5(1) = 5?
What if it were \sum_{n=0}^5{1}?
How about this one? \sum_{n=-5}^5{1}?

 

[SpaceDomain]

If the summation were
\sum_{n=1}^5{1}
can you see that it would be 1 + 1 + 1 + 1 + 1 = 5(1) = 5?
What if it were \sum_{n=0}^5{1}?
How about this one? \sum_{n=-5}^5{1}?

\sum_{n=1}^5{1} = 5

\sum_{n=0}^5{1} = 6

\sum_{n=-5}^{5}{1}

=\sum_{n=-5}^{-1}{1}

+ \sum_{n=1}^5{1}

+ 1 = 11

Okay. That helps a lot. Thank you all very much.