How do you compute the limit of {ln n - ln(n+1)}?

Answer 1

Use the laws of logs to combine these two terms:
ln n - ln(n+1) = ln(n/(n+1))

Now we'll look at the limit of n/(n+1). Dividing top and bottom by n, this becomes: 1/(1 + 1/n). Since 1/n->0 as n gets large,
1/(1 + 1/n) -> 1/(1+0) = 1

This means that ln(n/(n+1)) -> ln(1) = 0

So the limit of the sequence {ln n - ln(n+1)} is 0.

Answer 2

I assume you mean the limit as n approaches infinity. By logarithm rules, {ln(n)-ln(n+1)}=ln(n/(n+1))= ln((n+1-1)/(n+1)) = ln(1-1/(n+1)). As n approaches infinity, 1/(n+1) approaches zero, so ln(1-1/(n+1)) approaches ln(1)=0.

Answer 3

permit: L = lim (n-->infinity) n^{ln[(n + a million)/n]} ==> L = lim (n-->infinity) n^[ln(a million + a million/n)]. by skill of taking organic logarithms, we've: ln(L) = lim (n-->infinity) ln{n^[ln(a million + a million/n)]} ==> ln(L) = lim (n-->infinity) ln(a million + a million/n)*ln(n) ==> ln(L) = lim (n-->infinity) ln(a million + a million/n)/[a million/ln(n)]. by skill of L'Hopital's Rule: lim (n-->infinity) ln(a million + a million/n)/[a million/ln(n)] = lim (n-->infinity) [(-a million/n^2)/(a million + a million/n)]/{-a million/[n*ln^2(n)]} = lim (n-->infinity) [n*ln^2(n)]/(n^2 + n) = lim (n-->infinity) ln^2(n)/(n + a million) = 0. as a effect, ln(L) = 0 and: L = lim (n-->infinity) n^{ln[(n + a million)/n]} = e^0 = a million, as required. i'm hoping this helps!

Answer 4

As n goes to what? 0? Infinity?