# 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}}{2n^3-1} Question
Series 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.
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{2n^3-1}$$ 2021-01-14
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.

### Relevant Questions 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.
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^5}$$ Consider the series $$\sum_{n=1}^\infty\frac{(-1)^n}{n^2}$$
a) Show the series converges or diverges using the alternating series test.
b) Approximate the sum using the 4-th partial sum($$S_4$$) of the series.
c) Calculate the maximum error between partial sum($$S_4$$) and the sum of the series using the remainder term portion of the alternating series test. Consider the following convergent series.
a. Find an upper bound for the remainder in terms of n.
b. Find how many terms are needed to ensure that the remainder is less than $$10^{-3}$$.
c. Find lower and upper bounds (ln and Un, respectively) on the exact value of the series.
$$\sum_{k=1}^\infty\frac{1}{3^k}$$ Which of the series, and which diverge?
Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.)
$$\sum_{n=1}^\infty\frac{1}{2n-1}$$ Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series.
$$\sum_{n=2}^\infty\frac{(-1)^nn}{n^2-3}$$ Use the alternating series test to determine the convergence of the series
$$\sum_{n=1}^\infty(-1)^n\sin^2n$$ Use the Direct Comparison Test or the Limit Comparison Test to determine the convergence or divergence of the series.
$$\sum_{n=1}^\infty\frac{1}{\sqrt{n^3+2n}}$$ $$\sum_{n=1}^\infty\frac{\sin^2n}{n^3}$$ $$\sum_{n=1}^\infty\frac{(-1)^n}{n^5}$$ Determine whether the series $$\sum a_n$$ an converges or diverges: Use the Alternating Series Test.
$$\sum_{n=2}^\infty(-1)^n\frac{n}{\ln(n)}$$