# What is #lim_(n -> oo) n ln((n + 1)/n)#?

##### 1 Answer

The limit equals

#### Explanation:

Using logarithm laws:

#L = lim_(n -> oo) n(ln((n + 1)/n))#

#L = lim_(n->oo) n(ln(1 + 1/n))#

We can now do a little algebraic reworking:

#L = lim_(n-> oo) ln(1 + 1/n)/(1/n#

Now if you evaluate

#L = lim_(n-> oo) (-1/(n^2 + n))/(-1/n^2)#

#L = lim_(n->oo) n^2/(n^2 + n)#

#L = lim_(n-> oo) n^2/(n(n + 1))#

# L = lim_(n->oo) n/(n + 1)#

We now use partial fractions to decompose.

#n/(n + 1) = A/(n + 1) + B/1#

#n = A + B(n + 1)#

#n = A + Bn + B#

We can see instantly that

#L = lim_(n->oo) -1/(n + 1) + 1#

#L = lim_(n-> oo) -1/(n + 1) + lim_(n ->oo) 1#

The first limit is clearly

#L = 0 + 1#

#L = 1#

If we check graphically, we can see this is evidently true.

Hopefully this helps!