# Changing the running parameter

1. Nov 22, 2008

### Marin

hi there!

I have the following sum:

$$\displaystyle{\sum_{n=1}^\infty}(-1)^n\frac{(2z)^n}{(2n)!}$$

and I wanna let it run from n=0 to infty

it´s somehow obvious that:

$$\displaystyle{\sum_{n=0}^\infty}(-1)^n\frac{(2z)^n}{(2n)!} - 1=\displaystyle{\sum_{n=1}^\infty}(-1)^n\frac{(2z)^n}{(2n)!}$$

but how do I prove it? I tried to write n-1 instead n but when I calculated it it wasn´t in a way the result from the equation above.

Does anyone know how I should do this operation?

2. Nov 22, 2008

### HallsofIvy

The only difference between $\sum_{n=0}^\infty$ and $\sum_{n=1}^\infty$ is the n= 0 term! What is (-1)0 (2z)0/(2(0))! ? Now do you see why you need that "-1" outside the sum?

3. Nov 22, 2008

### Marin

yeah, yeah, I know why it has to be outside, maybe I have to ask my question more precisely. My qestion is, how to change the running parameter, what is the way one does this so that the -1 comes as result of this change?

4. Nov 22, 2008

### gabbagabbahey

Well, a summation is just a whole bunch of additions, so it should be obvious to you that

$$\sum_{n=0}^{\infty} a_n=a_0 + \sum_{n=1}^{\infty} a_n \Rightarrow \sum_{n=1}^{\infty} a_n= \sum_{n=0}^{\infty} a_n-a_0$$

That is really all there is to it. There is no need to change your running variable.

5. Nov 22, 2008

### Marin

:) ok, tnaks once again for the help