Alternative solutions to $\lim_{n\to\infty} \frac{1}{\sqrt{n}}\int_{ 1/{\sqrt{n}}}^{1}\frac{\ln(1+x)}{x^3}\mathrm{d}x$

Solution 1:

An alternate way is to notice that expanding the integrand also gives:

$$\frac{1}{x^2}-\frac{1}{x}<\frac{\ln(1+x)}{x^3}< \frac{1}{x^2}\;\,\text{for}\;\,0<x<1$$

Hence:

$$\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{1}{x^2}-\frac{1}{x}\,dx< \frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}dx< \frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{1}{x^2}dx$$

$$1-\frac{1}{\sqrt{n}}-\frac{\ln\sqrt{n}}{\sqrt{n}}<\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}dx< 1-\frac{1}{\sqrt{n}}$$

$$\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}\to 1$$

Solution 2:

By L`Hopital's rule:

$$ \lim_{n\rightarrow \infty} -\frac{\ln(1 + \sqrt{n}) n^{1.5} \frac{-1}{2 n^{1.5}}}{0.5 n^{-0.5}}$$ $$ = \lim_{n\rightarrow \infty} \ln(1 + \frac{1}{\sqrt{n}}) \sqrt{n}$$ Using the fact that $$\ln(1 + x) = x + O(x^2)$$

We see that the limit is $1$.

Solution 3:

Not strictly avoiding the integration, but expanding the integrand makes it pretty straightforward:

$$\frac{\ln(1+x)}{x^3}=\frac{1}{x^2}-\frac{1}{2x}+\frac{1}{3}-\frac{x}{4}+\cdots$$

$$\int_{\frac{1}{\sqrt{n}}}^1 \frac{\ln (1+x)}{x^3}dx=\left[-\frac{1}{x}-\frac{\ln x}{2}+\frac{x}{3}-\cdots\right]_{1/\sqrt{n}}^1=-\frac{3}{4}+\sqrt{n}+\frac{\ln \sqrt{n}}{2}-\frac{1}{3\sqrt{n}}+\cdots$$

$$\frac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1 \frac{\ln (1+x)}{x^3}dx=1-\frac{3}{4\sqrt{n}}+\frac{\ln \sqrt{n}}{2\sqrt{n}}-\frac{1}{3n}+\cdots\to 1$$

Solution 4:

$$\lim_{n\to \infty}\dfrac{1}{\sqrt{n}}\int_{\frac{1}{\sqrt{n}}}^1\dfrac{\ln(1+x)}{x^3}\mathrm dx=\lim _{t\to \infty}\frac{1}{t}\int_{\frac{1}{t}}^1\dfrac{\ln(1+x)}{x^3}\mathrm dx= \lim _{t\to \infty}\ln(1+t^{-1})t^3/t^2=1$$ By l'Hospital's rule and $\log(1+x)/x\to1$