Solving the limit of an integral Compute $$\lim _{n\to \infty }\left(\frac{\left(4n-4-2a_n\right)}{\pi }\right)^n$$ Where $$a_n=\int _1^n\:\frac{2x^2}{x^2+1}dx$$ The integral I solved and I got $a_n=2(x-\arctan(x))$ Afterwards, after I took into account the fact that it's a definite integral, I got $a_n=2(n-\arctan(n)-\frac{\left(4-\pi\right)}{4})$ I tried to meddle with the limit a bit too and I got $\lim _{n\to \infty }3^n$, which would be $\infty$. Is this correct? Since it seemed way too easy. Could someone check my answer?

$$\frac{2x^2}{1+x^2}=2-\frac2{1+x^2}\implies$$

$$a_n=\left.\int_1^n\frac{2x^2}{1+x^2}dx=\left(2x-2\arctan x\right)\right|_1^n=2n-2\arctan n-2+\frac\pi2\implies$$

$$\frac{4n-4-2a_n}\pi=\frac{4\arctan n}\pi-1\implies$$

$$\left(\frac{4n-4-2a_n}\pi\right)^n\xrightarrow[n\to\infty]{}...$$

For the last limit, you may want to use the exponential function:

$$\lim_{n\to\infty}n\log\left(\frac{4\arctan n}\pi-1\right)=\lim_{n\to\infty}\frac{\log\left(\frac{4\arctan n}\pi-1\right)}{\frac1n}\stackrel{l'H}=$$

$$=\lim_{n\to\infty}\frac{\frac\pi{4\arctan n-\pi}\frac4{\pi(1+n^2)}}{-\frac1{n^2}}=-\frac4\pi$$