How Accurate Is the Ninth Partial Sum of an Alternating Series?

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Homework Statement


if the series ∑(n=1, goes to infinity) (-1)^n/(n^3)
is approximated by its ninth partial sum, find a bound for truncation error

Homework Equations


Alternating Series Estimation Thm:
If alternating series is CONVERGENT, then truncation error for nth partial sum is less than U(sub(n+1)) and has the same sum as the first unused term:

|error|< U(sub(n+1))

The Attempt at a Solution


|error|< U(sub10)
U(sub10) = 1/10^3
|error|< 1/1000

is this right..?

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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I think that 'third partial sum' means
\sum_{n=1}^3 (-1)^n \frac{1}{n^3} = -1 + \frac{1}{2^3} - \frac{1}{3^3}.
 
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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