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

Dec 16, 2017

The limit equals $1$

#### Explanation:

Using logarithm laws:

$L = {\lim}_{n \to \infty} n \left(\ln \left(\frac{n + 1}{n}\right)\right)$

$L = {\lim}_{n \to \infty} n \left(\ln \left(1 + \frac{1}{n}\right)\right)$

We can now do a little algebraic reworking:

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

Now if you evaluate $n = \infty$ into both the numerator and the denominator, you will get $\frac{0}{0}$, so we can now apply l'Hospital's rule. By the chain rule, $\frac{d}{\mathrm{dn}} \ln \left(1 + \frac{1}{n}\right) = - \frac{1}{n} ^ 2 \cdot \frac{1}{1 + \frac{1}{n}} = - \frac{1}{{n}^{2} + n}$

$L = {\lim}_{n \to \infty} \frac{- \frac{1}{{n}^{2} + n}}{- \frac{1}{n} ^ 2}$

$L = {\lim}_{n \to \infty} {n}^{2} / \left({n}^{2} + n\right)$

$L = {\lim}_{n \to \infty} {n}^{2} / \left(n \left(n + 1\right)\right)$

$L = {\lim}_{n \to \infty} \frac{n}{n + 1}$

We now use partial fractions to decompose.

$\frac{n}{n + 1} = \frac{A}{n + 1} + \frac{B}{1}$

$n = A + B \left(n + 1\right)$

$n = A + B n + B$

We can see instantly that $B = 1$ and $A + B = 0$, therefore, $B = 1$ and $A = - 1$.

$L = {\lim}_{n \to \infty} - \frac{1}{n + 1} + 1$

$L = {\lim}_{n \to \infty} - \frac{1}{n + 1} + {\lim}_{n \to \infty} 1$

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.

Hopefully this helps!