1. **Problem Statement:** Examine the convergence of the series: (i) $$\sum_{n=1}^\infty \frac{(2n)! \cdot (5n)!}{(6n)!}$$ (ii) $$\sum_{n=1}^\infty \frac{\sin(n)}{n^2 + \pi n + 1} \cdot \frac{1}{2^n + 5}$$ Also, show that for any real $x > 0$, the series $$\sum_{n=1}^\infty \frac{x^n}{n^2}$$ converges to a real number. 2. **Raabe's Test (Κριτήριο Ρίχας):** Given a sequence $(a_n)$ of real numbers, suppose the limit $$\lambda = \lim_{n \to \infty} n \left(\frac{|a_n|}{|a_{n+1}|} - 1\right)$$ exists with $\lambda \in (-\infty, +\infty]$. - If $\lambda > 1$, then $\sum a_n$ converges absolutely. - If $\lambda < 1$, then $\sum a_n$ diverges. - If $\lambda = 1$, the test is inconclusive. 3. **Step-by-step solution:** **(i) Series with factorials:** - Let $a_n = \frac{(2n)! (5n)!}{(6n)!}$. - Compute the ratio: $$\frac{a_n}{a_{n+1}} = \frac{(2n)! (5n)! (6(n+1))!}{(6n)! (2(n+1))! (5(n+1))!}$$ - Using Stirling's approximation for large $n$: $$k! \sim \sqrt{2 \pi k} \left(\frac{k}{e}\right)^k$$ - Approximate each factorial and simplify the ratio inside the limit: $$\frac{a_n}{a_{n+1}} \approx \frac{(2n)^{2n} (5n)^{5n} (6n + 6)^{6n + 6}}{(6n)^{6n} (2n + 2)^{2n + 2} (5n + 5)^{5n + 5}} \times \text{(terms that cancel or tend to 1)}$$ - After simplification, the dominant terms give: $$\frac{a_n}{a_{n+1}} \approx \left(\frac{6n + 6}{6n}\right)^{6n} \times \left(\frac{2n}{2n + 2}\right)^{2n} \times \left(\frac{5n}{5n + 5}\right)^{5n} \times \text{constants}$$ - Using limits: $$\left(1 + \frac{1}{n}\right)^n \to e$$ - So, $$\left(\frac{6n + 6}{6n}\right)^{6n} = \left(1 + \frac{1}{n}\right)^{6n} \to e^6$$ $$\left(\frac{2n}{2n + 2}\right)^{2n} = \left(\frac{1}{1 + \frac{1}{n}}\right)^{2n} = \left(1 + \frac{1}{n}\right)^{-2n} \to e^{-2}$$ $$\left(\frac{5n}{5n + 5}\right)^{5n} = \left(1 + \frac{1}{n}\right)^{-5n} \to e^{-5}$$ - Multiplying these: $$e^6 \times e^{-2} \times e^{-5} = e^{-1}$$ - Therefore, $$\frac{a_n}{a_{n+1}} \to e^{-1}$$ - Now compute $\lambda$: $$\lambda = \lim_{n \to \infty} n \left(\frac{a_n}{a_{n+1}} - 1\right) = n (e^{-1} - 1)$$ - Since $e^{-1} - 1 < 0$, the limit tends to $-\infty$. - Hence, $\lambda < 1$ and by Raabe's test, the series diverges. **(ii) Series with sinc and rational terms:** - The general term is: $$a_n = \frac{\sin(n)}{n^2 + \pi n + 1} \cdot \frac{1}{2^n + 5}$$ - Note that $|\sin(n)| \leq 1$. - For large $n$, $n^2 + \pi n + 1 \sim n^2$ and $2^n + 5 \sim 2^n$. - So, $$|a_n| \leq \frac{1}{n^2} \cdot \frac{1}{2^n} = \frac{1}{n^2 2^n}$$ - Since $\sum \frac{1}{2^n}$ converges (geometric series) and $\frac{1}{n^2}$ is decreasing, the product converges absolutely by comparison test. - Therefore, the series converges absolutely. **(iii) Series $$\sum_{n=1}^\infty \frac{x^n}{n^2}$$ for $x > 0$:** - Since $n^2$ grows faster than $n$, and $x^n$ grows exponentially if $x > 1$, we consider cases: - If $0 < x < 1$, then $x^n \to 0$ exponentially, so the series converges absolutely. - If $x = 1$, the series is $$\sum \frac{1}{n^2}$$ which converges (p-series with $p=2 > 1$). - If $x > 1$, the terms $x^n / n^2$ grow without bound, so the series diverges. - The problem states $x > 0$ and the series converges to a real number, so it must be $0 < x \leq 1$. 4. **Final answers:** - (i) The series diverges. - (ii) The series converges absolutely. - (iii) For $0 < x \leq 1$, the series $$\sum_{n=1}^\infty \frac{x^n}{n^2}$$ converges to a real number.