Question

Test the series for convergence or divergence. [infinity] (−1)n + 1 3n5 n = 1 converges diverges Correct: Your answer is correct. If the series is convergent, use the Alternating Series Estimation Theorem to determine how many terms we need to add in order to find the sum with an error less than 0.00005. (If the quantity diverges, enter DIVERGES.) terms

Answer 1

Answer:

a.

The sequence that you are summing is decreasing and its limit is zero, therefore it converges.

b.

You need to sum at least 6 terms to get that precision

Step-by-step explanation:

You want to determine the convergence of the series

$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{3n^5}$

Since.

1.  $\frac{1}{3n^5}$ is a decreasing sequence

2. $\lim\limits_{n->\infty} \, \frac{1}{3n^5} = 0$

Therefore by the theorem of convergence for alternating series i is convergent.

To determine how many terms you need to add in order to find the sum with an error less than 0.00005 we use the estimation theorem for alternating series which states that for the n-error of an alternating series is upper bounded by the next term of the series. In math terms it looks like this.

Given

$s = \sum\limits_{n=0}^{\infty} (-1)^n a_n$

The error $R_n$ satisfies that

$R_n = |s - s_n| \leq a_{n+1}$

For this problem, we are looking for an $n$ such that

$\frac{1}{3n^5} \leq 0.00005$

then, solving the inequality you get that

$n \geq \sqrt[5]{\frac{1}{3*0.00005}} = 5.818$

Therefore you need to sum at least 6 terms to get that precision