Question

Test the following sequences for convergence or divergence. I will write the sequence

Answer 1

\[(e/ \pi) ^n \]
from n=1 to infinity

Answer 2

sorry it took solong...it lost connection

Answer 3

and that is a sequence

Answer 4

test if e/pi is < 1

Answer 5

it diverges because lim n->oo is not 0

Answer 6

a number less than 1 raised to a power gets smaller

Answer 7

so e/pi is less than 1

Answer 8

therefore (e/pi)^N --> 0 as N--> infinity

Answer 9

so it goes to 0 so it converges right?

Answer 10

yes

Answer 11

ok and now for the last one:

Answer 12

its a big expression so i will write it in parts

Answer 13

the numerator of the seq. is (-1)^(n-1)
the denominator of the seq is n(n+1)
from n=1 to infinity

Answer 14

It bounces back and forth around zero, but the denominator is approach infinity, while the numerator is either +1 or -1. So it converges on zero.

Answer 15

um how wud u write that using limits so that i can show a proof as my answer

Answer 16

\[\lim_{n \rightarrow \infty} \frac{(-1)^{n-1}}{n(n+1)}=0\]

Answer 17

Does your book have an example of how to write it?

Answer 18

i think that looks gud. it just shows it as:
If lim n->infinty of a sub n = L then a sub n converges

Answer 19

my question is does it converge to 0 because teh denominator apporaches 0 evn though numerator ggoes to +1/-1

Answer 20

I believe so. The test is | Sn - L| < epsilon for all n> N
where epsilon is as small as you wish.

Answer 21

That means once n gets big enough (bigger than N, whatever that happens to be), the distance from zero is less than epsilon.

Answer 22

Here on some notes
http://tutorial.math.lamar.edu/Classes/CalcII/Sequences.aspx

Answer 23

THANK U SOOOOO MUCH =) i appreciate all ur help

Answer 24

In the notes, see example 2c where he does an alternating sequence. The idea is if the absolute value of the sequence converges, then so does the alternating signed sequence.

Answer 25

good night

Answer 26

u too goodnite