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.

Alternating series remainder states that if a convergent alternating series satisfies the condition \(a_{n+1}\leq a_n\), then the absolute value of the remainder RN involved approximating the sum S by \(S_N\) is less than or equal to the first neglected term. That is, \(|S-S_N|=R_N\leq a_{N+1}\).

The given series is \(\sum_{n=1}^\infty\frac{(-1)^{n+1}}{2n^3-1}\),

Here \(a_n=\frac{1}{2n^3-1}\).

By the alternating series remainder theorem, \(|R_N|\leq a_{N+1}=\frac{1}{2(N+1)^3-1}\).

For margin of less than 0.001.

\(\frac{1}{2(N+1)^3-1}<0.001\)</span>

\(2(N+1)^3-1>\frac{1}{0.001}\)

\(2(N+1)^3-1>1000\)

\(2(N+1)^3>1001\)

\((N+1)^3>\frac{1001}{2}\)

\(N+1>(\frac{1001}{2})^{\frac{1}{3}}\)

\(N>6.639\)

That is, \(N\approx7\).

It appears that would need at least 7 terms.

Use of 7 terms, the sum is \(S=S_7=0.9474\) that has an error margin of less than 0.001

The required number of terms will be 7.

Alternating series remainder states that if a convergent alternating series satisfies the condition \(a_{n+1}\leq a_n\), then the absolute value of the remainder RN involved approximating the sum S by \(S_N\) is less than or equal to the first neglected term. That is, \(|S-S_N|=R_N\leq a_{N+1}\).

The given series is \(\sum_{n=1}^\infty\frac{(-1)^{n+1}}{2n^3-1}\),

Here \(a_n=\frac{1}{2n^3-1}\).

By the alternating series remainder theorem, \(|R_N|\leq a_{N+1}=\frac{1}{2(N+1)^3-1}\).

For margin of less than 0.001.

\(\frac{1}{2(N+1)^3-1}<0.001\)</span>

\(2(N+1)^3-1>\frac{1}{0.001}\)

\(2(N+1)^3-1>1000\)

\(2(N+1)^3>1001\)

\((N+1)^3>\frac{1001}{2}\)

\(N+1>(\frac{1001}{2})^{\frac{1}{3}}\)

\(N>6.639\)

That is, \(N\approx7\).

It appears that would need at least 7 terms.

Use of 7 terms, the sum is \(S=S_7=0.9474\) that has an error margin of less than 0.001

The required number of terms will be 7.