Question

use the alternating series test (if possible) to determine whether the series converges or diverges.

Answer 1

\[\sum_{n=1}^{\infty} \frac{ (-1)^n }{ \ln(n+12) }\]

Answer 2

@rational could you help me with a few more this time they are alternating series test

Answer 3

like i know an = 1/ln(n+12) and you get rid of the (-1)^n

Answer 4

its equal to zero so it converges right?

Answer 5

Yes but you need to show that it passes the requirements for applying alternating series test

Answer 6

http://gyazo.com/8ca143e34a12e6b8f6fa658c9ff877e7

Answer 7

1.
\[\lim\limits_{n\to\infty} \dfrac{1}{\ln(n+2)} = 0\]
so requirement 1 is met.


2.
show that the sequence \(\left\{\frac{1}{\ln(n+2)}\right\}\) is decreasing

Answer 8

\[\frac{ 1 }{ \ln(n+1+12) } < \frac{ 1 }{ \ln(n+12) }\]

Answer 9

is that how i show it?

Answer 10

that will do!

Answer 11

\[\ln(n+12+1) \gt \ln(n+12)~~ \implies~~ \frac{1}{\ln(n+12+1) } \lt \frac{1}{\ln(n+12)}\]

so the sequence is decreasing

Answer 12

we're good to use alternating series test as we have shown that the given series meets the hypothesis

Answer 13

Just conclude by saying that the series converges by alternating series test.

Answer 14

okay do you mind if we do two more examples?

Answer 15

Okay post

Answer 16

\[\sum_{n=1}^{\infty} (-1)^n \frac{ 5n^7 + 4 }{ 4n^6 + 3n + 5}\]

Answer 17

for this one i got an = \[\frac{ 5n^7 + 4 }{ 4n^6 + 3n +5}\]

Answer 18

and what do you do when the degree of the numerator is greater than the denominator

Answer 19

http://www.coolmath.com/precalculus-review-calculus-intro/precalculus-algebra/18-rational-functions-finding-horizontal-slant-asymptotes-01

Answer 20

when the degree of numerator is greater than the denominator, the graph wont settle... meaning, the terms keep growing/decreasing without any bound. so the limit is infinity

Answer 21

so if its infinity than checking if it decreases is not necessary?

Answer 22

The first requirement failed, so there is no point in testing second requirement.

we cannot use alternating series test, need to try some other test

Answer 23

what other test could we use?

Answer 24

could we use the limit test?

Answer 25

yes try it

Answer 26

alright

Answer 27

\[\lim_{n \rightarrow \infty} \frac{ 5n^7+4 }{ 4n^6 + 3n +5 }----->\lim_{n \rightarrow \infty} \frac{ 35n^6 }{ 24n^5 + 3 }-------->\frac{ 210n^5 }{120n^4 }\]

Answer 28

what happened to (-1)^n

Answer 29

i thought we didnt need it anymore since its not part of an

Answer 30

in "limit test" you need to take the entire expression.

Answer 31

thats not required only in alternatign series test

Answer 32

so it would be a negative value \[-\frac{ 35n^6 }{ 24n^5 + 3}\]

Answer 33

right?

Answer 34

how do you know whether (-1)^n is negative or positive ?

Answer 35

its not its indeterminate

Answer 36

Exactly!

Answer 37

look at it like this :

\[\lim_{n \rightarrow \infty} (-1)^n \frac{ 5n^7+4 }{ 4n^6 + 3n +5 } ~~=~~\left(\lim_{n \rightarrow \infty} (-1)^n\right) \cdot \left(\lim_{n \rightarrow \infty} \frac{ 5n^7+4 }{ 4n^6 + 3n +5 } \right)\]

Answer 38

you have worked earlier... the second limit is infinite, right ?

Answer 39

right

Answer 40

and the first limit seems to be toggling between +1 and -1

Answer 41

so clearly the overall limit also toggles bbetween +infty and -infty, yes ?

Answer 42

whatever it is, we're convinced it certainly doesn't converge yes ?

Answer 43

yes

Answer 44

So can we say the series diverges by "limit test"

Answer 45

yes we can but is it okay if i use another test when its asking to use the alternating series test

Answer 46

like do you think my teacher would give me credit if i use another test besides the one he ask to do

Answer 47

show ur work for alternating series test,
show that the first requirement of alternating series test is not met

Answer 48

then mention that you need to use another test as the hypothesis of "alternating series test" is not met

Answer 49

you need to show that all in work of this problem

Answer 50

oh yes of course

Answer 51

go through Example2 problem quick
http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx

Answer 52

oh i see similar to mine that problem is

Answer 53

so this other one is \[\sum_{n=1}^{\infty} \frac{ (-1)^n * [\sin (3n)]^2}{ n }\]

Answer 54

i got an= \[\frac{ [\sin(3n)]^2 }{ n }\]

Answer 55

Yes test the first condition

http://gyazo.com/8ca143e34a12e6b8f6fa658c9ff877e7

Answer 56

\[\lim_{n \rightarrow \infty} \frac{ [\sin(3( \infty ))]^2 }{ \infty } = 0\]

Answer 57

what about second condition

Answer 58

\[\frac{ [\sin(3(n+1))]^2 }{ n+1 } < \frac{ [\sin(3n)]^2 }{ n }\]

Answer 59

it is not true i plug in 2 for n and i got a higher number on the left side

Answer 60

yeah strictly speaking, it fails the second condition

Answer 61

so it is divergent?

Answer 62

lets look at its graph maybe

Answer 63

alright

Answer 64

Keep in mind, alternating series test is used only to show convergence.

failure of conditions in alternating series test doesn't imply divergence.