1. **Problem statement:** Consider the sequence $u_n = n\sqrt{n} = n^{3/2}$. We want to understand its divergence behavior and apply the definition of divergence to $u_n \in (10^6, +\infty)$. Also, prove that $\lim_{n \to +\infty} \ln(\ln(2^n + 1)) = +\infty$. 2. **Definition of divergence to $+\infty$:** A sequence $(u_n)$ diverges to $+\infty$ if for every $A > 0$, there exists $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$, $u_n > A$. 3. **Apply to $u_n = n^{3/2}$ with $u_n \in (10^6, +\infty)$:** We want to find $A$ and $n_0$ such that $u_n > A$ for all $n \geq n_0$ and $u_n > 10^6$. Set $A = 10^6$. Solve $n^{3/2} > 10^6$: $$n^{3/2} > 10^6 \implies n > (10^6)^{2/3} = 10^{4}$$ So, $n_0 = 10^4$. Thus, for all $n \geq 10^4$, $u_n > 10^6$. 4. **Proof that $\lim_{n \to +\infty} \ln(\ln(2^n + 1)) = +\infty$:** - Note that $2^n$ grows exponentially. - Then $\ln(2^n + 1) \sim \ln(2^n) = n \ln 2$ for large $n$. - So $\ln(\ln(2^n + 1)) \sim \ln(n \ln 2) = \ln n + \ln(\ln 2)$. - Since $\ln n \to +\infty$ as $n \to +\infty$, the limit is $+\infty$. 5. **Divergence of sequences:** 1) $u_n = \cos\left(\frac{n\pi}{4}\right)$: - The sequence takes values in a finite set $\{\cos(k\pi/4) : k=0,1,...,7\}$ and is periodic. - It does not converge, so it diverges. 2) $u_n = \left(1 + \frac{1}{n}\right)^{n^2}$: - Rewrite as $\exp\left(n^2 \ln\left(1 + \frac{1}{n}\right)\right)$. - For large $n$, $\ln\left(1 + \frac{1}{n}\right) \sim \frac{1}{n}$. - So exponent $\sim n^2 \cdot \frac{1}{n} = n \to +\infty$. - Hence $u_n \to +\infty$, diverges. 3) $u_n = \frac{1 + n(-1)^n}{n}$: - For even $n$, $u_n = \frac{1 + n}{n} = 1 + \frac{1}{n} \to 1$. - For odd $n$, $u_n = \frac{1 - n}{n} = 1 - 1 = 0$. - The subsequences have different limits, so $u_n$ diverges. 4) $u_n = \frac{n! + 3^n}{n - 2^n}$: - For large $n$, $2^n$ dominates denominator negatively. - Numerator grows roughly like $n!$ which grows faster than $3^n$. - Since denominator $\sim -2^n$, the ratio behaves like $\frac{n!}{-2^n}$. - $n!$ grows faster than any exponential, so $|u_n| \to +\infty$. - Hence $u_n$ diverges. **Final answers:** - For $u_n = n^{3/2}$, $A = 10^6$, $n_0 = 10^4$. - $\lim_{n \to +\infty} \ln(\ln(2^n + 1)) = +\infty$. - The sequences in II) are divergent by the reasons above.