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Convergence of series

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I am pretty new with this serie I am supposed to find convergent or divergent.
\(\displaystyle \sum_{k=1}^\infty[\ln(1+\frac{1}{k})]\)

progress:
\(\displaystyle \sum_{k=1}^\infty[\ln(1+\frac{1}{k})]= \sum_{k=1}^\infty [\ln(\frac{k+1}{k})] = \sum_{k=1}^\infty[\ln(k+1)-\ln(k)]\) so we got that
\(\displaystyle \lim_{n->\infty}(\ln(2)-\ln(1))+\)\(\displaystyle (\ln(3)-\ln(2))+...+(\ln(n+1)-\ln(n))\)
and this is where I am stuck
:confused:

Regards,
\(\displaystyle |\pi\rangle\)
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Re: serie

You've done fine up until this point. Consider the first few terms you have written :

$$\ln(2) - \ln(1) + \ln(3) - \ln(2) + \cdots$$

Do you see that $\ln(2)$ cancels out? Can you, similarly, show that some further terms cancel outs too? If so, can you determine what is finally left?

Keyword : Telescoping series
 
Last edited:

Petrus

Well-known member
Feb 21, 2013
739
Re: serie

You've done fine up until this point. Consider the first few terms you have written :

$$\ln(2) - \ln(1) + \ln(3) - \ln(2) + \cdots$$

Do you see that $\ln(2)$ cancels out? Can you, similarly, show that some further terms cancel outs too? If so, can you determine what is finally left?

Keyword : Telescoping series


EDIT : This can be shown in another way though rather easily by noting that $\log (1 + 1/k) \geq \log(1/k)$.
Ohh I see it was infront of my eyes... I did not even checking those will cancel out...! Thanks alot now I got it!
We got left
\(\displaystyle \lim_{n->\infty}\ln(n+1)-\ln(1)=\infty\)
Thanks for taking your time! Guess I need to wake up!:p

Regards,
\(\displaystyle |\pi\rangle\)