Question

series

Answer 1

\[\sum_{n=3}^{\infty} \frac{ (-1)^{n-1} }{ nlnn }\]

I got converges by Alternating series test, but am not sure.

Answer 2

@ganeshie8

Answer 3

Finding the derivative gives a decreasing function, taking limit n-> infinity = 0

Answer 4

Seems fine to me :) Lim as a_{n} goes to infinity is 0 and you can show it's decreasing with the first derivative test. So yes, converges.

Answer 5

All the other tests would be inconclusive

Answer 6

Yeah and \[b_n+1 < b_n\] I think

Answer 7

Yeah, which is what the first derivative test shows, so you're right.

Answer 8

Alright, cool :)! Just wanted to make sure it's convergent and wolfram gives http://www.wolframalpha.com/input/?i=series+n%3D3+to+infinity%28-1%29%5E%28n%29%2F%28nlnn%29 haha so I wasn't sure.

Answer 9

n n-1 what ever the alternating is, should still be convergent

Answer 10

u an rewrite (-1)^(n-1) as cos npix

Answer 11

\[\sum_{n=3}^{\infty} \frac{ (-1)^{n-1} }{ n\ln n } = \sum_{n=3}^{\infty} \frac{ \cos((n+1)\pi) }{ n\ln n }\]

next what dan

Answer 12

i dunno lol i was gonna show that as a continuous function is it converging

Answer 13

one way to see its converging for sure is that its n ln n so its >1 and

Answer 14

u just got -1 and +1 so

Answer 15

1/n^k, k>1 and -1/n^k, K>1

Answer 16

are both converging series

Answer 17

Oh nice using a p series?

Answer 18

I was thinking about \[\sum_{n=1}^{\infty} \frac{ 1 }{ nlnn }\] which would be divering using the comparison test haha

Answer 19

diverging*

Answer 20

im not sure if this is valid though why cant we just say
n ln n = n^k, k>1

Answer 21

as a series

Answer 22

is thhaat wrong??

Answer 23

we are starting at n=0 so we never worry about ln being negative

Answer 24

p series requires k >=2

Answer 25

so n ln n must > n, and and sum of series 1/n^K, where K>1 is convergin

Answer 26

oh oh really

Answer 27

T_T

Answer 28

nlnn happens to be the border case

Answer 29

n < nlnn < n^2

Answer 30

p series requires >1 I thought

Answer 31

right right !

Answer 32

where did i get this idea about border case hmm

Answer 33

http://prntscr.com/537vy4

Answer 34

yeah k>1 is sufficient and harmonic series is the true border line ;)

Answer 35

yup

Answer 36

Thanks guys ^.^

Answer 37

lol