Happy New Year: Infinite Sum Question Explanation

[matematikuvol]

Happy new year. All the best.

I have one question. Is it true?
\sum^{\infty}_{k=0}a_kx^k=\sum^n_{k=0}a_{n-k}x^{n-k}
I saw in one book relation
\sum^{\infty}_{k=0}\frac{(2k)!}{2^{2k}(k!)^2}(2xt-t^2)^k=\sum^{n}_{k=0}\frac{(2(n-k))!}{2^{2(n-k)}((n-k)!)^2}(2xt-t^2)^{n-k}
Can you give me some explanation for this step?

 

[tiny-tim]

hi matematikuvol!

the RHS is a function of n (ie, it's different for different n), but the LHS isn't …

so that equation can't possibly be correct

perhaps they mean
\sum^{n}_{k=0}a_kx^k=\sum^n_{k=0}a_{n-k}x^{n-k}
(which obviously is true)

 

[oli4]

Hi matematikuvol,
Hi tiny-tim
I think what is missing in your equations is the term limit.
That is, the infinite sum would be (might be) the limit of the other sum when n goes to infinity.
But in this case, beware that although what tiny-tim says about it being obviously true (you just reverse the summation order) this is not necessarily true anymore when you take the limit.
For it to hold even when n goes to infinity your sum must be absolutely convergent, otherwise changing terms order in interesting ways could lead you to find the limit to be just about anything (0, ∏, whatever)

 

[JG89]

Hi matematikuvol,
Hi tiny-tim
I think what is missing in your equations is the term limit.
That is, the infinite sum would be (might be) the limit of the other sum when n goes to infinity.
But in this case, beware that although what tiny-tim says about it being obviously true (you just reverse the summation order) this is not necessarily true anymore when you take the limit.
For it to hold even when n goes to infinity your sum must be absolutely convergent, otherwise changing terms order in interesting ways could lead you to find the limit to be just about anything (0, ∏, whatever)

This particular derangement doesn't affect the sum when passing to the limit because \forall n \in \mathbb{N} , \sum_{k=0}^n a_k x^k = \sum_{k=0}^n a_{n-k} x^{n-k}. Thus \sum_{k=0}^{\infty} a_k x^k = \sum_{k=0}^{\infty} a_{n-k} x^{n-k}

 

[mathman]

This particular derangement doesn't affect the sum when passing to the limit because \forall n \in \mathbb{N} , \sum_{k=0}^n a_k x^k = \sum_{k=0}^n a_{n-k} x^{n-k}. Thus \sum_{k=0}^{\infty} a_k x^k = \sum_{k=0}^{\infty} a_{n-k} x^{n-k}

Your last remark is incorrect, since you are still looking at an n in each term of the infinite sum.
You would need ∞-k, which is meaningless in this context.