Hello,
I am looking up different logarithmic tutorials but cannot find which law justifies this:
$\displaystyle ln(1+\frac{1}{n}) = -ln(1-\frac{1}{n+1})$
could $\displaystyle ln x \leq (x-1)$ be of any help?
Thank you in advance!
Hello,
I am looking up different logarithmic tutorials but cannot find which law justifies this:
$\displaystyle ln(1+\frac{1}{n}) = -ln(1-\frac{1}{n+1})$
could $\displaystyle ln x \leq (x-1)$ be of any help?
Thank you in advance!
Write:
$\displaystyle \ln \left(1+\frac{1}{n}\right)=-\ln \left(1-\frac{1}{n+1}\right)$ as
$\displaystyle \ln \left(\frac{n+1}{n}\right)=-\ln \left(\frac{n}{n+1}\right)$
And we know (in general): $\displaystyle y\log_a(x)=\log_a(x^y)$, so ...