f:(0;+∞)-R,f(x)=ln(1+1/x)
Calculate the limit of the string:

Question

f:(0;+∞)->R,f(x)=ln(1+1/x)
Calculate the limit of the string:

anonymous · Answer

$$(An)n \ge1 ;An=f(1)+f(2)+...+f(n)-\ln(2n+1)$$

ganeshie8 · Answer

telescoping series

ganeshie8 · Answer

$\large \sum \limits_{x=1}^n f(x)  $


$\large \sum \limits_{x=1}^n \ln(1 + \frac{1}{x}) $

$\large \sum \limits_{x=1}^n \ln(\frac{x+1}{x}) $


$\large \sum \limits_{x=1}^n \ln(x+1) - \ln (x) $


$\large [\ln(1+1) - \ln (1)  ]  +  [\ln(2+1) - \ln (2)  ]  + ...  [\ln(n+1) - \ln (n)  ] $


$\large [\ln(2) - \ln (1)  ]  +  [\ln(3) - \ln (2)  ]  + ...  [\ln(n+1) - \ln (n)  ] $

ganeshie8 · Answer

everything in the middle cancels out, leaving u wid :

$\large   -\ln(1) +\ln(n+1)$

anonymous · Answer

an this row tends to....................?

ganeshie8 · Answer

since ln(1) = 0,

that gives u   $\large   \ln(n+1)$

anonymous · Answer

yeah i got it thanks

ganeshie8 · Answer

so,  


$\sum \limits_{x=1}^n \ln(1 + \frac{1}{x}) =  \ln(n+1) $

ganeshie8 · Answer

$An=f(1)+f(2)+...+f(n)-\ln(2n+1)$


$An=\ln(n+1)-\ln(2n+1)$


$An=\ln(\frac{n+1}{2n+1})$

ganeshie8 · Answer

u wlc :)