## anonymous 5 years ago Power Series: sigma ((-1)^n x^n)/(n+1) as n-> infinity

1. anonymous

$\sum_{0}^{\infty} ((-1)^n x^n)/(n+1)$

2. anonymous

i know that im suppose to replace all n's with n+1 and then use ration test but im not sure what steps to take next

3. anonymous

Hi donutboy, how are you? Do you know the series representation for ln(1-x)?

4. anonymous

You want to check the convergence of the series or you want to find its sum?

5. anonymous

ratio* hi sklee, im fine, no i dont know the ln(1-x)

6. anonymous

im looking for the radius of convergence and interval of convergence

7. anonymous

ok great, if you want to find the radius and interval of convergence, then you dont have to know that series representation

8. anonymous

For radius of convergence, i assume you will use the Ration test, and not using the formula. So I will show you the work using the test.

9. anonymous

Let a_n = (-1)^n x^n / (n+1), then by the Ratio test, lim |a_(n+1)/a_n| = lim |[x^(n+1)/(n+2)] [(n+1)/(x^n)] = lim |x {(n+1)/(n+2)}| Because lim (n+1)/(n+2) = 1 and the series converges if the value of the limit of |a_(n+1)/a_n| is less than 1, we get |x|<1

10. anonymous

ahh icic thank you so much! my prof. goes way to fast and its hard to keep up

11. anonymous

So the radius of convergence is 1, and the interval we are considering is -1 < x < 1. So we need to check the convergence of the series at the end point x =-1 and x =1

12. anonymous

at x = -1, the series become \sum {1/(n+1)}, which is divergent as it is a variation of harmonic series. at x=1, the series becomes \sum {(-1)^n / (n+1)}, which is a variation of the Alternating harmonic series, and this series converges. So the interval of convergence is -1 < x =< 1

13. anonymous

quesiton on the lim |a_(n+1)/a_n| = lim |[x^(n+1)/(n+2)] [(n+1)/(x^n)] = lim |x {(n+1)/(n+2)}| part what happens during lim |[x^(n+1)/(n+2)] [(n+1)/(x^n)] ?

14. anonymous

x^(n+1)/x^n can be simplified to become x

15. anonymous

and i just arrange the term (n+1) and (n+2) as a fraction

16. anonymous

oh okay thank you again

17. anonymous

you're welcome

18. anonymous

$\sum_{0}^{\infty} (-1)^n (x^(2n)/ (2n!)$

19. anonymous

so far ive gotten to $\lim_{n \rightarrow \infty} \left| (2n!(x^(2n+1)/x^(2n)(2n+2) \right|$

20. anonymous

ok, very good

21. anonymous

now the [x^(2n+1)]/x^(2n) is simplified to x

22. anonymous

and (2n)! / (2n+2)! is simplified to 1/[(2n+1)(2n+2)] because (2n+2)! = (2n+2)(2n+1)[(2n)!]

23. anonymous

alrighty

24. anonymous

the limit of 1/[(2n+1)(2n+2)] is 0, so the series converges for all x

25. anonymous

so the radius is infinity and the interval of convergence is (-infinity, infinity)

26. anonymous

yup, you are right!

27. anonymous

thanks! im starting to get this now!

28. anonymous

great, you can do it donutboy. All the best!

29. anonymous

thanks ^__^ youre awesome!