Alternating Series

[Ashlander]

Hey everyone! I just registered and would like to say hello to the community, but I also am very stuck with this calculus homework and also need some help regarding alternating series, if anyone could please help!

Consider the series:

(E n=1 to infinity) ((-1^n)sin(n)) / (n^6 +1)
(E n=1 to infinity) ((-1)^n) / ((n!)^2)

a. for each of these, decide whether the alternating series test are satisfied
b. for those series satisfying the conditions, decide how many terms need to be added up in order to reach within 10^-8 of the sum of the series. Give a decimal approximation of the sum of one of the series with maximum allowed error of 10^-8.

I have absolutely no idea how to progress with this question and I am really stumped, if anyone could please help me! The (E n=1 to infinity) means Sigma Notation where the sum taken should be from 1 to infinity. Also, part b. just numbs my brain :shock:

 

[stapel]

a. for each of these, decide whether the alternating series test are satisfied....

I have absolutely no idea how to progress with this question....

Does your book not provide a statement of the Alternating Series Test? (If it had, then determining whether the requirements of the Test are satisfied would be a simple matter of comparing the series with the stated conditions, is why I ask.)

Thank you.

Eliz.

 

[Ashlander]

Yeah, it does, but the problem is the example problems in the textbook are very simple compared to the above homework problems, I've tried a few methods but I have no idea whether or not I'm doing it right since I just get stuck in the middle... I'm so lost

 

[galactus]

Are the conditions of the alternating test met?.

\(\displaystyle a_{1}>a_{2}>a_{3}>......>a_{k}>....\)

and

\(\displaystyle \lim_{k\to\infty}a_{k}=0\)


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

\(\displaystyle \frac{1}{(n!)^{2}}>\frac{1}{((n+1)!)^{2}}\)....check.

\(\displaystyle \displaystyle\lim_{n\to\infty}\frac{1}{(n!)^{2}}=0\)....check.

The conditions are met. It must converge.

As a matter of fact, it converges to -0.7761092209

 

[Ashlander]

hey that helped a lot!! Thanks very much :lol: