# Thread: Calculus 4: Alternating Series

1. ## Calculus 4: Alternating Series

1) How to determine the convergence or divergence of the following series:

a) lim of ((-1)^(n+1) *n)/ (2n-1) when n=1

b) lim of ((-1)^(n+1) *n^2)/ (n^2 + 5) when n=1

c) lim of (2(-1)^(n+1))/ (e^n + e^-n)= lim of ((-1)^(n+1) * sech n) when n=1

2) Determine whether the following series converges conditionally or absolutely, or diverges.

a) lim of ((-1)^(n+1))/ (n+1) when n=1

b) lim of ((-1)^n * e^(-n^2)) when n=0

c) lim of ((-1)^(n+1) * arctan n) when n=1

3) Using the alternating Series Remainder Theorem, determine the number of terms required to approximate the sum of the following series with an error of less than 0.001.

a) lim of ((-1)^(n+1)) / (n^2) when n=1

2. Use the test for Divergence

Theorem:

if $\displaystyle \lim_{n\to\infty}a_n\ne0$ or does not exits the series

$\displaystyle \sum_{n=1}^\infty a_n$ is divergent

This should help with a few. you should look up AST AST estimation theorem.

3. I don't understand could you please expand on the explanation?

4. Originally Posted by googoogaga
I don't understand could you please expand on the explanation?
What don't you understand? The Theorem? Please be clear.

$\displaystyle \sum_{n=1}^\infty\frac{(-1)^{n+1}n}{2n-1}$

so

$\displaystyle a_n=\frac{(-1)^{n+1}n}{2n-1}$

So divide both the numerator and denominator by n

$\displaystyle a_n=\frac{(-1)^{n+1}}{2-\frac{1}{n}}$

as n goes to infinity the SEQUENCE of $\displaystyle a_n$ diverges.

so by the Divergence Theorem the SERIES divierges

5. Thank you Empty Set. the first part makes sense now that you've broken it down for me. The other ones (2& 3) I am not so sure about...