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

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SUMMARY

The discussion centers on estimating the truncation error of the ninth partial sum of the convergent alternating series ∑(n=1, ∞) (-1)^n/(n^3). According to the Alternating Series Estimation Theorem, the truncation error is bounded by the absolute value of the next term, U(sub(10)), which is calculated as 1/10^3. Thus, the truncation error is definitively less than 1/1000, confirming the accuracy of the approximation using the ninth partial sum.

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syeh
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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}.
 

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