Alternating Series: Evaluate E(n = 1) to Infinity

vigintitres
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Homework Statement



E(n = 1) to infinity ((-1)^n+1)/n^6


Homework Equations


This needs to be in the proper form with the exponent on an being n - 1 not n + 1


The Attempt at a Solution



I don't know how to get the problem into the proper for to evaluate it as an alternating series
 
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I don't know what you mean by "standard form". If you're just trying to prove that the sum converges, the alternating series test requires that you show

(1) a_n is strictly alternating (i.e. a_{n+1}=-a_n)

(2) \lim_{n \to \infty} a_n = 0

(3) |a_{n+1}| \leq |a_n|

EDIT: You could also use direct comparison to show that the sum is absolutely convergent.
 
I don't see why you need a "proper form". I mean it is alternating signs after all. If you wanted the exponent to be n-1 you could just replace n+1 with n-1 since you're just dividing by (-1)^2 = 1. But there really is no point in doing that.

Use the absolute convergence test first when dealing with alternating series. You'll see that in some cases such as this one, using it determines convergence/divergence easily.
 
But if you really must have "proper form", (-1)n+1= (-1)n-1+ 2= (-1)n+1(-1)2= (-1)n-1 because (-1)2= 1.
 
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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