## Math2400 one year ago radius of convergence and the interval of convergence for the given power series

1. Math2400

$\sum_{n=1}^{\infty} \frac{ n ^{2}x ^{2n} }{ n! }$

2. Math2400

my radius is 1. And it think the interval is (-1,1) Can anyone verify if that's right?

3. Zarkon

what test did you use?

4. Math2400

ratio

5. Zarkon

do it again

6. Math2400

ok

7. Math2400

wait can the radius be infinity??

8. Zarkon

yes

9. Math2400

ok..and the interval is (-infin, infin) ?

10. Zarkon

yes

11. Math2400

lol kk thanks. that's what i got first but i wasn't sure if that was possible haha

12. xapproachesinfinity

i got oo as well :)

13. Math2400

perfect thanks!! :)

14. xapproachesinfinity

limit (x^2 . (n+1/n)^2/(n+1))=0

15. Math2400

how about this one if u don't mind: $\sum_{n=1}^{\infty} \frac{ n^{e} }{ e ^{n} }$ i'm pretty sure the radius is 1 right?

16. xapproachesinfinity

that says for any value of x rho is always zero

17. Math2400

and then interval is (2,4) or maybe (2,4] not sure, i suck at these

18. xapproachesinfinity

i would do root test with that one

19. xapproachesinfinity

did you try ratio?

20. Zarkon

there is no x in your last problem

21. Math2400

yes(: I got the radius to be 1 and then i know the interval is (2,4) but I'm not sure

22. xapproachesinfinity

eh hold on you are asking for convergence because there is not x

23. xapproachesinfinity

you need x

24. Math2400

ohh wait sorry i posted the wrong one. That one is a converging series haha. I meant to post this one: $\sum_{n=1}^{\infty} \frac{ (x-3)^n }{ \sqrt{n} }$

25. xapproachesinfinity

oh ok i would go with root test with that too limit n^1/n=1 rho=|x-3|<1 so raduis is 1 -1<x-3<1 ===> 2<x<4

26. xapproachesinfinity

i guess you got this :)

27. Math2400

ok so radius is 1 and then did u check the end points?

28. xapproachesinfinity

No i didn't x=4 you have series 1/n^1/2 diverges x=2 (-1)^n/rootn i believe this converge by alternating series test

29. Math2400

so would it be (2,4] or just (2,4) that's what I'm confused on

30. xapproachesinfinity

no converge on 2 not on 4 so [2,4) otherwise 2<=x<4

31. Math2400

ohh haha kk got it :) Thank you so much!

32. xapproachesinfinity

no problem :)

33. xapproachesinfinity

actually you seem to me good at this, just few more practice and you master it

34. Math2400

haha thank you, means a lot @xapproachesinfinity