<s>[INST] Evaluating $\lim\limits_{n \to \infty} \int_0^1 \frac{\ln^n (1+x)}{1+x^2}dx$ Can you please verify my proof below?
My solution
Notice that $f(x):=\ln^n(x)$ and $g(x):=\dfrac{1}{1+x^2}$ is continuous over $[0,1]$, and $g(x)>0$. As per the first integral mean value theorem, on can have
$$0\leq \int_0^1 \frac{\ln^n (1+x)}{1+x^2}dx=\ln^n(1+\xi)\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}\ln^n(1+\xi)\leq \frac{\pi}{4}\ln^n(1+1)\to 0,$$
which implies $$\lim\limits_{n \to \infty} \int_0^1 \frac{\ln^n (1+x)}{1+x^2}dx=0.$$
 [/INST] It's correct, but you can get the same bound easier, without using MVT:
For $0\le x \le 1$ we have $0 \le \ln(1+x) \le \ln 2$, so $0\le \frac{\ln^n(1+x)}{1+x^2} \le \frac{\ln^n 2}{1+x^2}$ and
$$ 0\le \int_0^1 \frac{\ln^n(1+x)}{1+x^2} dx \le \int_0^1  \frac{\ln^n 2}{1+x^2} dx=\frac{\pi}{4} \ln^n 2$$
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