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

1 Answer
Noah G
Dec 16, 2017

The limit equals #1#

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 #n = oo# into both the numerator and the denominator, you will get #0/0#, so we can now apply l'Hospital's rule. By the chain rule, #d/(dn) ln(1 + 1/n) = -1/n^2 * 1/(1 + 1/n) = -1/(n^2 + n)#

#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 #B = 1# and #A + B = 0#, therefore, #B = 1# and #A = -1#.

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

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

The first limit is clearly #0# and the second #1#.

#L = 0 + 1#

#L = 1#

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

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Hopefully this helps!

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