Quote Originally Posted by akhayoon View Post
$\displaystyle \sum\frac{lnk}{e^{k}}$

so the book is telling me to use the ratio test...

and you get$\displaystyle \frac{ln(k+1)}{elnk}=\frac{1}{e}$

so now I guess how would ln(k+1)/lnk go to 1 would be my basic question...

all I know is is that ln(n)/n goes to 0 as n goes to infinity but i haven't seem something like this before..
if you are asking how $\displaystyle \lim_{x \to {\infty}}\frac{ln(x+1)}{ln(x)}=1$? it is either logical conisdering with huge numbers +1 doesnt matter or applying l'hopitals we get $\displaystyle \lim_{x \to {\infty}}\frac{\frac{1}{x+1}}{\frac{1}{x}}=\lim_{x \to {\infty}}\frac{x}{x+1}$ and either knowing the leading coefficent trick or applying l'hopital's again we get $\displaystyle \lim_{x \to {\infty}}\frac{1+0}{1}=1$