Is \sum_5^\infty \frac{1}{(n-4)^2} = \sum_1^\infty \frac{1}{n^2} Legal?

[e^(i Pi)+1=0]

I just wanted to check that this was legal.

$\sum_5^\infty \frac{1}{(n-4)^2} = \sum_1^\infty \frac{1}{n^2}$ ?

 

[Dick]

Sure it is. They are the same series, aren't they?

 

[e^(i Pi)+1=0]

Thanks

 

[HallsofIvy]

A slightly more formal derivation would be:
$$\sum_{n=5}^\infty \frac{1}{(n-4)^2}$$
Let i= n- 4. Then $(n-4)^2= i^3$ and when n= 5, i= 5- 4= 1. Of course, i= n-4 goes to infinity as n goes to infinity so
$$\sum_{n=5}^\infty \frac{1}{(n-4)^2}= \sum_{i=1}^\infty \frac{1}{i^2}$$

But both "n" and "i" are "dummy" variables- the final sum does not involve either- so we can change them at will. Changing "i" to "n" in the last sum,
$$\sum_{n=5}^\infty \frac{1}{(n-4)^2}= \sum_{i=n}^\infty \frac{1}{n^2}$$

The crucial point is that, as Dick said, "they are the same sequence":
$$\sum_{n=5}^\infty \frac{1}{(n-4)^2}+ \frac{1}{(5-4)^2}+ \frac{1}{(6-4)^2}+ \frac{1}{(7- 4)^2}+ \cdot\cdot\cdot= 1+ \frac{1}{4}+ \frac{1}{9}+ \cdot\cdot\cdot$$
$$\sum_{n= 1}^\infty \frac{1}{n^2}= \frac{1}{1^2}+ \frac{1}{2^2}+ \frac{1}{3^2}+ \cdot\cdot\cdot= 1+ \frac{1}{4}+ \frac{1}{9}+ \cdot\cdot\cdot$$