Question

calculus. did i do this right?

Answer 1

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

Answer 2

=\[\frac{ d }{ dx }\lim_{n \rightarrow \infty}\frac{ \frac{ 1 }{ n } }{ n }*\frac{ n }{ n }=\frac{ 1 }{ n^2 }\]

Answer 3

then by the p series 2>1 so the series is convergent?

Answer 4

i always screw up the ln stuff. just cant seem to get it down

Answer 5

you are taking the derivative of top and bottom, as I see?
but isn't (-1)^(n-1) also on the top ?

the top is not just infinity, it is going from negative to positive infinity depending on whether n-1 is even or odd.

Answer 6

oh yes... so how should i have done it with this then?

Answer 7

if i do the alternating series test... how is ln (n+1) less than ln (n)?

Answer 8

i didnt mean to say "how is" i just mean "is"

Answer 9

http://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+n%3Dinfinity+%28%28%28-1%29%5E%28n-1%29+times+%28ln+n%29%29%2Fn%29

Answer 10

this is what wolf got

Answer 11

not much, hah ?

Answer 12

were you trying to use alternating test
and then use losptial to find the limit of ln(n)/n as n approaches infinity ?

Answer 13

nope. they didnt tell us anything. i do know the answer is convergent. I just dont know how to get there properly

Answer 14

freckles! yes, that is what i was attempting

Answer 15

\[\lim_{n \rightarrow \infty}\frac{\ln(n)}{n}=\lim_{n \rightarrow \infty}\frac{(\ln(n))'}{(n)'}=?\]

Answer 16

after you show that approaches 0
you need to show ln(n)/n is decreasing if possible

Answer 17

i thought it was \[\frac{ \frac{ 1 }{ n } }{ n }\] but then i realized i didnt take the bottom properly and now i think its just \[\frac{ 1 }{ n }\]. which is divergent by the harmonic series but convergent by the Alt Ser Test, i think

Answer 18

I am not sure if ln(n+1) is less than ln(n)

Answer 19

i want to say yes it is, but the ln's always mess me up in my head

Answer 20

\[a\gt b \implies \ln(a) \gt \ln(b)\]

Answer 21

http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx
I'm going by this site
we have the first thing we need
now we need to see if we have the second thing:

\[\lim_{n \rightarrow \infty}\frac{\ln(n)}{n}=\lim_{n \rightarrow \infty}\frac{1}{n}=0\]

we need to see if :
\[\frac{\ln(n)}{n} > \frac{\ln(n+1)}{n+1} \text{ for all } n>0\]

Answer 22

on my calc, it says that ln(1)=0 and ln(2) = .693147..

Answer 23

calculator*

Answer 24

that is true since a>b implies lna>lnb :)

Answer 25

hi X! what do you mean by that ? is this a rule i dont know about?

Answer 26

hey there :)
no i was just clarifying when you said ln1=0 and ln2=0.6
i said in general a>b gives lna>lnb this is always true

Answer 27

why don't we do the integral here?

Answer 28

it is the alternating series test

Answer 29

i need to prove that a_n > a_n+1 but i dont see it..im doing something wrong

Answer 30

if you feel the integral will work, id be happy to try it with you

Answer 31

i hate that one the most tho lol

Answer 32

@xapproachesinfinity . after applying the integral test, that just ends with 1/n, right? and by theory, that is 0= convergent. is this all i need?

Answer 33

you may use first derivative to prove lnx/x is decreasing for x > e

Answer 34

f(x) = lnx/x

f'(x) = (1-lnx)/x^2 < 0 for all x > e

Answer 35

@ganeshie8 becaue e=1 right?

Answer 36

do you mean ln e = 1 ?

Answer 37

yes, sorry

Answer 38

hmm now that i think about integral test may not be your way out
i was just throwing ideas lol

Answer 39

im taking any suggestion, dont apologize. Its appreciated

Answer 40

hmm so lnx/x > ln(x+1)/x+1 for x>e

Answer 41

ok, so let me sum this up\[\sum_{n=1}^{\infty}(-1)^{n-1}\frac{ \ln(n) }{ n }\rightarrow \lim_{n \rightarrow \infty}\frac{ \ln(n)' }{ n' }= \frac{ 1 }{ n }\]
which is zero and then \[a_n >a_n+1 \] because ln(1)<lne......?

Answer 42

then these two factors meet the requirements of the alternating series test for convergence?

Answer 43

hold up
what exactly is your goal here ?

Answer 44

to prove A) that the lim n>infinty= zero and that a_n > a_n+1

Answer 45

you had some feeling that the given infinite sum converges to some number as it might satisfy the requirements of alternating series test

Answer 46

You need to prove 3 things to conclude the series covnerges using alternating series test :

1) \(\frac{\ln n}{n}\ge 0\) for all \(n\)

2) \(\lim\limits_{n\to \infty}\frac{\ln n}{n} = 0\)

3) \(\{\frac{\ln n}{n}\}\) is a decreasing sequence

Answer 47

ok..

Answer 48

Did you prove each of them ?

Answer 49

i proved the lim=0 and you tell me i am aloud to use lne=1 to prove decreasing... but does this also prove that a_n>a_n+1 ? this is where i was before you told me about the lne

Answer 50

a_n>a_n+1 is the same as showing that the series decrease

Answer 51

considering that a_n=lnn/n

Answer 52

oh... ok. so what i posted above... where i said "ok, so lets sum this up"... is that correct?

Answer 53

the limit has been proved to be zero
you need 1)
i think the third step has also been proved but for x>e

Answer 54

the notation is not so great in that reply. you might be right but your prof will give you 0 marks if you submit that

http://gyazo.com/d89d7df15ca83f70e4756bd4320e18f1

Answer 55

oh yeah the way you put it is not good :P lol

Answer 56

lol. how should i put it?

Answer 57

\[\lim_{n\to \infty} \ln n/n=\lim_{n\to \infty }\frac{1}{n}=0\]
and you don't need lne>ln1 thingy

Answer 58

^

Answer 59

thanks. its just kinda confusing to me that they want me to prove all these things and then just telling them the lim=0 is enough.

Answer 60

hey do we need to show that it is decreasing for any positive n?

Answer 61

Alt Series Test, from my instructions says to prove lim=o and that a_n> a_n+1

Answer 62

that will do for second requirement

next, you may use first derivative test to prove that the sequence is decreasing.
after that you will be ready to conclude the series converges using alternating series test

Answer 63

my personal response would be the argument that\[\frac{ 1 }{ n }>\frac{ 1 }{ n+1 }\]

Answer 64

Yes. I am also trying to work it without using derivatives :

\[n < n+1 \implies \dfrac{1}{n} \gt \dfrac{1}{n+1}\]

I'm bit stuck after this.. not sure how to proceed :/

Answer 65

if i can use that... then we have met the requirements

Answer 66

use what ?

Answer 67

derivatives ?

Answer 68

the argument that 1/n> 1/n+1 to satisfy that an is bigger

Answer 69

1/n > 1/(n+1) is true and we can use it

but this itself doesn't prove the sequence is decreasing

Answer 70

this is where i get confused, because i thought we proved that with the ln e> x

Answer 71

@ganeshie8 didn't you prove the series decreasing using derivatives?
why you trying other alternatives ?
i was thinking about ln n/n >0 for any n
is this necessary

Answer 72

also, i understand your reasoning. but my instructions do not ask me to prove decreasing. although proving an>an+1 kinda does all by itself, right?

Answer 73

yes that is the same thing

Answer 74

ok. I think i will stick with what we have here then. I TOTALLY appreciate all the help guys. wish i knew how to give more than one medal

Answer 75

there is a way to give more than one medal
but its more of like a bug... I'll give medals to everyone that helped in this thread :)

Answer 76

btw sometimes you're not allowed to use derivatives because sequences and series are covered before derivatives and integrals

(save integral test)

Answer 77

Ok! thank you!

Answer 78

hmm we didn't do them yet! this is part of calc2
the last thing we deal with is series and sequence in the program

Answer 79

i'm in calc2 still in by parts lol
we are slow hahaha

Answer 80

i hate by parts! and i am gonna have to brush back up on that at the the end of this semester because our final is comprehensive

Answer 81

hehehe last two weeks i was doing some interesting by parts integrals

Answer 82

i bet

Answer 83

look through my questions i have posted, you will see all the integration by parts questions i posted

Answer 84

a lot of your question are about series lol
actually i like series xD i just need to refresh things

Answer 85

but if you look way back into lets say beginning of january, you will see the integration stuff

Answer 86

i see, i will take a look later
gotta go see ya

Answer 87

ok have good one