Question

Sequences:

Answer 1

Converges or Diverges?

\[\LARGE \frac{(-1)^n}{2\sqrt{n}}\]

Answer 2

alternating sequences converge if...
1. Its terms decrease in magnitude
2. The terms converge to 0.

Answer 3

have you tried the ratio test?

Answer 4

Alternating series test[edit]
This is also known as the Leibniz criterion. If \(\sum_{n=1}^\infty a_n\) is a series whose terms alternative from positive to negative, and if the limit as n approaches infinity of a_n is zero and the absolute value of each term is less than the absolute value of the previous term, then \(\sum_{n=1}^\infty a_n \) is convergent.

source : wikipedia

Answer 5

try solving

Answer 6

\[\LARGE \lim_{n
\rightarrow \infty}~\frac{(-1)^n}{2\sqrt{n}}\]
As a limit?

Answer 7

how else laughing out loud

Answer 8

You'd be surprised >_>

Answer 9

Wouldn't that make it an indeterminate at the top and infinity at the bottom?

Answer 10

try what hartnn suggested

Answer 11

i think its conv

Answer 12

do know leibnitz theorem for a alternating series?

Answer 13

it is converge but heres what i do

Answer 14

Yup, each terms gets smaller and smaller.. \(\Large |-1/2|, |1/2\sqrt{2}|, |-1/2\sqrt{3}|, |1/4|,... \)

Answer 15

I'm technically not suppose to know any "theorems" by this chapter..

Answer 16

you don't

Answer 17

Or section

Answer 18

just use whatever else you know

Answer 19

when n is odd lim =0
when its even lim =0

Answer 20

a series of the form
\[un=\sum_{n=1}^{\infty}(-1)^{n}un\] converges if
1){un} is monotonic decreasing and
2)\[\lim_{n \rightarrow \infty}un=0\]

Answer 21

check if the conditions satisfies

Answer 22

its not monotonic....
u can devide it into seq
you would have one decrease and one increase

Answer 23

it is convergent the conditions are satisfying

Answer 24

just devide it into two subsequences

\(\LARGE \lim_{n
\rightarrow \infty}~\frac{ -1 }{2\sqrt{2n+1}}\)
\(\LARGE \lim_{n
\rightarrow \infty}~\frac{1 }{2\sqrt{2n}}\)

if lim 1=lim 2 = C
s.t C\(\neq 0\)
then its conv.

Answer 25

i made a typo sry
c\(\neq \infty \)

Answer 26

\[u_n-u_{n-1}>0\]
and also
\[\lim_{n \rightarrow \infty}\frac{ (-1)^n }{ 2\sqrt(n)}=0\]