Question

Series

Answer 1

geometric or arithmetic \(\huge\color{black}{???}\)

Answer 2

\[\sum_{n=1}^{\infty}\ln(2^n+1)-\ln(2^n)\]

Answer 3

@ganeshie8 @Abhisar @Callisto @Compassionate @uri @Kainui @eliassaab @inkyvoyd
@Joel_the_boss @Coolsector @Amistre64 @TheSmartOne @bibby @geerky42 @confluxepic @inkyvoyd
@Nnesha @camerondoherty @ParthKohli @Zale101 @poopsiedoodle

Answer 4

Find the nature of this Series !

Answer 5

@perl @Zarkon

Answer 6

i remember you posting a similar question few days ago
this has to do with telescoping right ?

Answer 7

Those are not related

Answer 8

plus I don't think you can telescope an exp form

Answer 9

are you trying to find the value of series or just want to know whether it converges ?

Answer 10

studying the converging

Answer 11

I'll be back in 15 mins till then you can post your ideas

Answer 12

my ideas are to try ratio test / comparison test
have you tried them yet ?

Answer 13

An observation:\[\ln(2^n+1)-\ln(2^n)=\ln\frac{2^n+1}{2^n}=\ln\left(1+\frac{1}{2^n}\right)<-n\ln2\]

Answer 14

Greater than*, sorry!

Answer 15

Hmm this would suggest that the given series diverges, but it doesn't... interesting

Answer 16

siths how about comparing with geometric series 1/2^n ?

Answer 17

Yup that'll work. Just wondering why what I said doesn't...

Answer 18

\[\lim\limits_{n\to \infty} \frac{\ln (1+\frac{1}{2^n})}{\frac{1}{2^n} }= 1\]

since the series SUM 1/2^n converges, SUM ln(1+1/2^n) also converges by lmit comparison test

Answer 19

so this question proves wolfram is not reliable?

Answer 20

lol why wolfram says this series converges right ?

Answer 21

one sec let me check again..

Answer 22

http://www.wolframalpha.com/input/?i=sum%28log_e%282%5En%2B1%29-log_e%282%5En%29%2Ci%3D1...infinity%29
it says by limit test series diverges

Answer 23

it says it converges when ln(2^n)=ln(1+2^n)
I don't know the point in it saying that though

Answer 24

wut...
http://www.wolframalpha.com/input/?i=Sum%5BLog%5B2%5En%2B1%5D-Log%5B2%5En%5D%2C%7Bn%2C1%2CInfinity%7D%5D

Answer 25

interesting
here it says it converges
http://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cln%281%2F2%5En%2B1%29

Answer 26

lol i'm an idiot

Answer 27

i put i=1..infinity
and used n instead

Answer 28

sorry guys

Answer 29

so yes it converges by the wolfram test :p

Answer 30

wolfram test haha! for a moment i doubted the goddess wolfram !!

Answer 31

Lol converges by the wolfram test, I gotta remember that.

Answer 32

The comparison that @SithsAndGiggles was illegitimate, \[\ \ln\left(1+\frac{1}{2^n}\right)>-n\ln2\]

This statement is wrong because the left term tends towards 0 while the right term actually tends towards negative infinity. So although he is not wrong in writing this inequality, there is more than one way to diverge ;P

Answer 33

Oh I see, we have to keep end behavior in mind as \(n\to\infty\)

Answer 34

The series converges to 0.868877
See
http://www.wolframalpha.com/input/?i=sum+%28log%28+2%5En%2B1%29+-log%282%5En%29%2C+n%3D1+to+infinity