Question

Series

Answer 1

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

@ganeshie8

Is this a telescoping series?

I did test for divergence got 0 which is inconclusive but it's not right.

Answer 2

try comparison test with harmonic series

Answer 3

first show that ln(1 + 1/n) > 1/n for all n > 1

Answer 4

Limit comparison right?

Answer 5

yes anything will do

Answer 6

So let \[a_n= \ln \left( 1+\frac{ 1 }{ n } \right)\]

bn will be the inside then? or just 1/n

Answer 7

\[b_n = \frac{ 1 }{ n }\]

Answer 8

The ln function is greater than 1/n is it not

Answer 9

you want the known diverging series to be less, so try \(\large b_n = \dfrac{0.5}{n}\)

Answer 10

So the constant doesn't matter?

Answer 11

You can make it anything you want

Answer 12

multiplying a constant by a diverging series won't make it converging
so it doesn't matter

Answer 13

Ah right

Answer 14

if a series is diverging, then so is the series multiplied by a non zero constant

Answer 15

or you could even compare it with ln(1/n)

\[\large b_n = \ln(\frac{1}{n})\]

Answer 16

\[\lim_{n \rightarrow \infty} \frac{ a_n }{ b_n } = \frac{ \ln \left( 1+\frac{ 1 }{ n } \right) }{ \frac{ 0.5 }{ n } }\] \[= \lim_{n \rightarrow \infty} \ln \left( 1+\frac{ 1 }{ n } \right)n\] actually what if I let that be 1/n then I can just jump to the expression above.. and divide everything by n and get it to diverge? But then I'll be back to step 1 haha.

Answer 17

Yeah I think that would be better

Answer 18

yeah showing \(\large \ln \left(1+\frac{1}{n}\right) \gt \ln\left(\frac{1}{n}\right)\) is trivial

Answer 19

also since ln is strictly increasing function and since \(\large i \gt j \implies \ln(i) \gt \ln(j)\), the series \(\ln (\frac{1}{n})\) diverges too. Having said this, u can use this as \(b_n\) in your comparison test

Answer 20

if everything above doesn't look straightforward, you could take partial sums manually and see where the series is heading

Answer 21

if u proved ln(1+1/n) > ln ( 1/n) ( any way u would like to show )
and took 1/n as a reference which is div
the comparison ends hmm

Answer 22

see why i like comparison , lol

Answer 23

yeah thats the reason i like comparison test

Answer 24

Oh same pinch !

Answer 25

So overall would this series diverge?

Answer 26

it has to diverge because the series with terms lesser term values diverges :

\(\large \ln(1/n)\) diverges \(\implies \ln(1+1/n)\) diverges by comparison test because \(\large\ln(1+1/n) \gt \ln(1/n)\)

Answer 27

That makes sense, thanks dude!

Answer 28

each term in ln(1+1/n) series is greater than the corresponding term in ln(1/n) series

so the series in question grows faster than the diverging series

Answer 29

keep this in ur mind if u have two series
\(y_n \), \(x_n \) then comparison test is :-

1_ if \(x_n>y_n\) and \(y_n\) diverge then \( x_n \) diverge
2_ if \(x_n >y_n\) and \( x_n \) converge then \( y_n\) converge

Answer 30

Yup, I just read my notes and used it, thanks :)

Answer 31

btw , he is not a dude :P

Answer 32

but he's a he

Answer 33

This is the first time I've called him dude, sounds weird, I'll keep it professional next time.

Answer 34

lol u got scared bhahahaha just messing with u :P

Answer 35

Lol, I don't want to disrespect ganeshie :p

Answer 36

Thanks again for the help both of you :D

Answer 37

ur wlc dude xD

Answer 38

lol dude is not a disrespectful/weird word duede

Answer 39

hey you can't compare it with ln(1/n) as the terms are not positive
use something else if you want it done using comparison test

Answer 40

why ?

Answer 41

comparison test requires terms in both series to be > 0, right ?

Answer 42

ok yess

Answer 43

oh i got it hmmm

Answer 44

can we use -ln(1/n) ?

Answer 45

i guess , since ln(1+1/n) >0

Answer 46

nvm it wont work

Answer 47

why , it would work :O

Answer 48

-ln(1/n) = ln(n) which grows faster than ln(1+1/n)

so not useful

Answer 49

mmmm

Answer 50

actually his original idea was straightforward :

\[\large \begin{align} \sum_{n=1}^{\infty} \ln \left( 1+\frac{ 1 }{ n } \right) &= \sum_{n=1}^{\infty} \ln \left( n+1 \right)-\ln(n) \\~\\&= \lim\limits_{k\to \infty } \ln(k+1) \\~\\&= \text{diverges}\end{align} \]

Answer 51

telescoping would have been sufficient :)

Answer 52

haha xD

Answer 53

@ganeshie8 Can you please show the telescoping step by step haha? I have an idea but I just want to see it.