Question

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

Answer 1

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.

Hopefully this helps!