Equivalence of Two Infinite Series

Bashyboy
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I am having a difficult time seeing how \sum_{n=0}^{\infty} ((-1)^n + 1)x^n is equivalent to 2\sum_{n=0}^{\infty} x^{2n}
 
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What is the difficulty?

take a look at ##((-1)^n +1)##, if n is even, what is it? If n is odd, what is it?
or just write out the first 5 or so non-zero terms of each and compare.
 
Oh, I see. So, any odd power would give a trivial answer, and we would disregard those?
 
Oh, I see. So, any odd power would give a trivial answer, and we would disregard those?
It is not "trivial" it is zero. Zero terms do not change the sum so it does not matter if you regard them them or not.

Sometimes the notation can hide stuff if you are not used to it - when stuck, try writing out a bunch of terms.
 
Thank you for your help.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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