Evaluate $\lim\limits_{n \to \infty} n \ln\frac{a(\sqrt{n}-1)}{a(\sqrt{n}-1)+b}$

Question

For $0<a,b<1$, evaluate $$\lim\limits_{n \to \infty} n \ln\frac{a(\sqrt{n}-1)}{a(\sqrt{n}-1)+b}$$

I was thinking of using the Taylor expansion of $\ln(1-x)$ but it did not give me a any bounds.

Answer 1

Let $x=\sqrt{n} - 1$ so that $x\to\infty$ and the limit is transformed into $$\lim_{x\to\infty} (1+x)^2\log\left(1-\frac{b}{ax+b}\right)$$ which can be rewritten as $$\lim_{x\to\infty} (1+x)\left(-\frac{b(1+x)}{ax+b}\right)\dfrac{\log\left(1-\dfrac {b} {ax+b} \right) }{\dfrac{-b} {ax+b}} $$ The last factor tends to $1$ (because $b/(ax+b) \to 0$), the middle one tends $-b/a$ and the first factor tends to $\infty $ so the desired limit is $-\infty$.

The above makes use of the standard limit $$\lim_{t\to 0}\frac{\log (1+t)}{t}=1$$

Answer 2

Note that: $$\lim\limits_{n \to \infty} n \ln\frac{a(\sqrt{n}-1)}{a(\sqrt{n}-1)+b}=\lim\limits_{n \to \infty} \ln\frac{1}{\left(1+\frac{b}{a(\sqrt{n}-1)}\right)^n}=\\ \ln \frac{1}{\lim\limits_{n \to \infty} \left(1+\frac{b}{a(\sqrt{n}-1)}\right)^{\frac{a(\sqrt{n}-1)}{b}\cdot \frac{bn}{a(\sqrt{n}-1)}}}=\ln \frac1{e^{+\infty}}=-\infty.\\$$

Answer 3

You can use the Taylor Series expansion like you suggested. You can rewrite the limit as

$\lim_{n \to \infty} n \log\left(1 - \frac{b}{a \sqrt{n-1} + b}\right) $.

You can then take the second order Taylor expansion:

$n (- \frac{b}{a \sqrt{n-1} + b} - (\frac{-b}{a \sqrt{n-1} + b})^2)$ + higher order terms.

Rewriting gives

$- \frac{b}{n}{a \sqrt{n-1} + b} - \frac{b^2 n}{(a \sqrt{n-1} + b)^2}$ + higher order terms.

We know from the properties of Taylor series that the last term in the expansion bounds all of the remaining terms. The first term goes to $-\infty$ and the second term is finite. Consequently, the limit is $-\infty$.