## anonymous 4 years ago i need to find the interval of convergence of the series (from 1 to infinity) of ((-1^n+1)(x-2)^n)/n2^n and then check for the convergence at the endpoints. i got the interval but i am having trouble with checking the endpoints. help ):

1. anonymous

$\sum_1^{\infty}\frac{(-1)^{n+1}(x-2)^n}{n2^n}$

2. anonymous

i am going to make a guess that the radius is 2, so that you will have $|x-2|<2$ is that what you got?

3. anonymous

YES. and then i added 2 to both sides. so 0<x<4 ?

4. anonymous

that is what i got but i kind of guessed. ok so now we have to check at x = 0 and x = 4 right?

5. anonymous

yerp....

6. anonymous

maybe i am off by a minus sign, not sure, but this is an alternating series so surely converges as the terms go to zero, so we can include 0

7. anonymous

wait. how did you do that. how does it cancel off?

8. anonymous

i cancled the $2^n$

9. anonymous

oooh lol sorry

10. anonymous

my mistake i meant $\sum\frac{(-1)^n}{n}$ oops

11. anonymous

i canceled wrong...

12. anonymous

no actually my algebra is terrible, can you explain the cancelling please?

13. anonymous

but conclusion is right. converges because you have an alternating series wehre terms go to zero

14. anonymous

you have in the numerator $(-2)^n$ and in the denominator you have $2^n$ but $(-2)^n=(-1)^n2^n$ so you can cancel a $2^n$top and bottom

15. anonymous

kick the minus sign in to the first part to make it $(-1)^n$ instead of $(-1)^{n+1}$

16. anonymous

that just leaves you with $\sum\frac{(-1)^n}{n}$

17. anonymous

ohh okay. and then by alternating series test it converges. got it! what about at x=4?

18. anonymous

well i didn't do that yet, so lets try it

19. anonymous

haha okay.

20. anonymous

i think this one is a problem because it will not alternate

21. anonymous

i think it will alternate? and it diverges because 1/n is harmonic.

22. anonymous

thank you so much though! do you know how to differentiate and integrate the series ones too?

23. anonymous

wait wait!!

24. anonymous

hold on a second i think i might have made a big big mistake!!

25. anonymous

why? i think it is correct...

26. anonymous

let me be more careful with zero

27. anonymous

whoa did i make a mistake!! lets go slow

28. anonymous

oh oops i think the x=4 is convergent too.

29. anonymous

lets look at the numerator when x = 0

30. anonymous

$(-1)^{n+1}(-2)^n$ $(-1)^{n+1}(-1)^n2^n$ $(-1)^{2n+1}2^n$

31. anonymous

now we can certainly cancel a $2^n$ with the denominator

32. anonymous

get $\sum\frac{(-1)^{2n+1}}{n}$

33. anonymous

ohh i seee...

34. anonymous

ratio test?

35. anonymous

but $2n+1$ is odd, and so the numerator is -1

36. anonymous

in other words you get $\sum\frac{-1}{n}$ which diverges for sure

37. anonymous

ohhh okay okay. sounds good. um can you help me with another one?

38. anonymous

sorry about that , it was a big mistake on my part. i just thought to kick over the minus sign, but i forgot to add the exponents

39. anonymous

we did not do it with 4 yet, wanna try it?

40. anonymous

4 is going to converge, because this time the numerator is $(-1)^{n+1}2^n$

41. anonymous

cancel the $2^n$ get $\sum \frac{(-1)^{n+1}}{n}$

42. anonymous

and this one alternates, so it converges for x = 4 but not for x = 0

43. anonymous

oh okay i get it!!

44. anonymous

yeah well i made a mistake, sorry it took so long

45. anonymous

you were so much help regardless. i barely know what i am doing. can i ask you for help on another one?

46. anonymous

sure but post in a new thread and i will come look it is much easier than scrolling down

47. anonymous

thanks !

48. anonymous

yw