Question

determine convergence or divergence using appropriate test, identify test. infnity over E over n=2 (((-1)^n)/(n ln n))

Answer 1

\[\sum_{n=2}^{\infty} f(n)\]

Answer 2

\[\sum_{n=2}^{\infty}((-1)^{n}/n \times \ln(n))\] i suppose, converges with leibnits crit

Answer 3

leibnitz*

Answer 4

i do not know this

Answer 5

\[\frac{(-1)^n}{n\ ln (n)}\]it flips back and forth from - to + right?

Answer 6

but does it converge between them ... hmmm

Answer 7

sure terms go to zero and it alternates

Answer 8

ln(inf) is a large number; so yes, i agree. it goes to zero while alternating

Answer 9

large number

Answer 10

http://www.wolframalpha.com/input/?i=sum+%28%28%28-1%29^n%29%2F%28n+ln+n%29%29%2C+n+%3D+2+to+infinity

Answer 11

i get it is an alternating series , but what test would i use to tell if conv or diverging

Answer 12

can we |restrain it| inside an absolute value?

Answer 13

doesnt converge with absolutes

Answer 14

can we try integral test?

Answer 15

integral can be done, but is unnecesary

Answer 16

i cant recall too many more tests; ratio test? comparison to familiar stuff tests?

Answer 17

\[\sum\frac{-1^n}{n\ \sqrt{n}}\] converges

Answer 18

this one is not absolutely convergent.

Answer 19

ratio,root tests

Answer 20

right, since the absolutes diverge :)

Answer 21

for alternating series the leibnitz test is used, , it states that for \[\sum_{a}^{\infty}(f(x))\] if \[\lim_{x \rightarrow \infty}(f(x))\] converges, then (-1)^n*f(x) also converges

Answer 22

made a little mistake, a (-1)^n and n=a should be written insidde the sum

Answer 23

thank you everyone