## blackstreet23 one year ago Find the radius and the interval of convergence sigma (X^k)/(k+1)

1. amistre64

first take the ratio, and work the limit

2. blackstreet23

$\sum_{k=0}^{\infty} \frac{ X^k }{ k+1 }$

3. blackstreet23

I did that well. I am having troubles testing the limits

4. amistre64

hmm, show your work, lets see how well you got to that point then

5. blackstreet23

I dont understand how to test if it decreasing in the alternating series test

6. amistre64

so your trying to test for x=-1 ?

7. blackstreet23

give me a second

8. amistre64

$\lim_{k\to\infty}\frac{x^{n+1}/(k+2)}{x^n/(k+1)}$ $\lim_{k\to\infty}x\frac{k+1}{k+2}$ $|x|\lim_{k\to\infty}\frac{k+1}{k+2}=|x|$

9. blackstreet23

10. amistre64

when x=1, this is a harmonic series ... when x=-1, its an alternating harmonic series ... do you recall the convergence of those?

11. blackstreet23

harmonic no

12. blackstreet23

how do you do it?

13. amistre64

the harmonic does not converge, the alternating does ...

14. blackstreet23

but why? Could you explain?

15. blackstreet23

what is a harmonic series? and how do i know if it converges or diverges?

16. amistre64

harmonic is$\frac11+\frac12+\frac13+\frac14+\frac15+...$ the proofing is textbook, but it eludes me

17. blackstreet23

so is like $\frac{ 1 }{ K! }$

18. amistre64
19. blackstreet23

ohh my bad $\frac{ 1 }{ k }$

20. blackstreet23

ohh ok i see if it has a K on the bottom

21. beginnersmind

@amistre "the proofing is textbook, but it eludes me" 1/3 + 1/4 > 1/2 1/5 + 1/6 + 1/7 + 1/8 > 1/2 1/9 + ... 1/16 > 1/2 etc.

22. blackstreet23

I still dont get how can someone recognize that a function is a harmonic series just by looking at it

23. amistre64

the alternating harmonic series converges :) that proof eludes me as well ... just recall it from the classes.

24. blackstreet23

is because of the K?

25. amistre64

1/n is the harmonic series by definition

26. amistre64

yes

27. blackstreet23

so K by itself would be a harmonic series?

28. blackstreet23

2K +1

29. amistre64

1/k^p converges for p > 1

30. blackstreet23

10 k -2

31. blackstreet23

isnt it 1/k^p a p-series?

32. amistre64

yes

33. blackstreet23

ok sooo if it has a K is a harmonic series and if it has an exponent is a p series?

34. amistre64

1/k = 1/k^1 by definition, 1/k is the harmonic series.

35. amistre64

that is why the p series test says that if p > 1 it converges, but if 0<p<= 1 it diverges

36. blackstreet23

so it diverges?

37. amistre64

why do i feel like this is just going in circles ....

38. blackstreet23

and please special help in the alternating series one !

39. blackstreet23

sorry i am just reinforcing the knowledge you are giving me by enphasing what you say lol

40. amistre64

the alternating harmonic series converges, by properties that i cant recall either. they are explained better by the texts

41. amistre64
42. amistre64

radius = 1 interval is: [-1,1)

43. blackstreet23

Ohh ok. I am so sleepy right now haha. I will analyse it more tomorrow :). Thanks a lot ! anything I will send you more messages :D

44. amistre64

im getting tired as well, but there are plenty of smarter people then me about :)