Given that


I have to find the limit

$$\lim_{n\to \infty} n(\sqrt[n^2]{nI_n}-1)$$

I tried to use $0 < \text{tg} x < 1$ for $0<x<\pi/4$ and i found that $0<I_n < \frac{3}{n}\left(e^{n\pi/4}-1\right)$. I think with this, the limit should be $0$, but I don't know.

  • $\begingroup$ Welcome to Math.SE! Interesting idea, but with $n\to \infty$ you have $e^{nx}/n \to \infty$ as well $\endgroup$ – gt6989b Feb 25 '20 at 22:22

Split $I_n=I_1+I_2$, where:

$$I_1=\int_0^{\frac{\pi}{4}}e^{nx}(\tan^{n-1}x+\tan^{n+1}x)\,dx,\ \ \ I_2=\int_0^{\frac{\pi}{4}}e^{nx}(\tan^{n}x)\, dx$$

Now, integrating by parts:

$$\begin{aligned} I_1&=\int_0^{\frac{\pi}{4}}e^{nx}\tan^{n-1}x (1+\tan^2x)\,dx\\ &= \int_0^{\frac{\pi}{4}}e^{nx}\tan^{n-1}x\cdot (\tan x)'\,dx\\ &= \int_0^{\frac{\pi}{4}}e^{nx}\frac{1}{n}(\tan^n x)'\,dx\\ &= \frac{1}{n}e^{nx}\tan^nx\bigg|_0^{\frac{\pi}{4}} - \frac{1}{n}\int_0^{\frac{\pi}{4}}ne^{nx}(\tan^{n}x)\, dx\\ &=\frac{1}{n} e^{\frac{n\pi}{4}}-I_2 \end{aligned} $$

Thus $I_n=\dfrac{1}{n} e^{\frac{n\pi}{4}}$, and the limit is:

$$\begin{aligned} \lim_{n\to \infty} n(\sqrt[n^2]{nI_n}-1) &= \lim_{n\to \infty}n(e^{\frac{\pi}{4n}}-1) \\ &= \frac{\pi}{4} \cdot \lim_{n\to \infty}\frac{e^{\frac{\pi}{4n}}-1}{\frac{\pi}{4n}}\\ &=\frac{\pi}{4}\\ \end{aligned}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy