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# Use Theorem Alternating Series remainder to determine the number of terms required to approximate the sum of the series with an error of less than 0.001. sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^5} # Use Theorem Alternating Series remainder to determine the number of terms required to approximate the sum of the series with an error of less than 0.001. sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^5}

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Series asked 2020-12-30
Use Theorem Alternating Series remainder to determine the number of terms required to approximate the sum of the series with an error of less than 0.001.
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^5}$$

## Answers (1) 2020-12-31
Given that
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^5}$$
We need use theorem alternating series to determine the number of terms requires to approximate the sum of the series with an error of less than 0.001.
Given series is alternate series with $$a_n=\frac{1}{n^5}$$
We know that
$$|R_n|\leq a_{N+1}=\frac{1}{(N+1)^5}$$
for an error less than 0.001N must satisfy the inequality.
$$\frac{1}{(N+1)^5}<0.001$$</span>
$$\frac{1}{0.001}<(N+1)^51000<(N+1)^5$$</span> taking $$\frac{1}{5}$$ root on both sides
$$1000^{\frac{1}{5}} \(3.98 \(2.98 So we will need at least 4 terms to get an error less than 0.001 ### Relevant Questions asked 2021-01-13 Use Theorem Alternating Series remainder to determine the number of terms required to approximate the sum of the series with an error of less than 0.001. \(\sum_{n=1}^\infty\frac{(-1)^{n+1}}{2n^3-1}$$ asked 2021-03-07
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