4.1. Find the first 2 terms of the sequence, \( \left(a_{n}\right) \) where \( a_{n+1}=\frac{9^{n+1}}{10^{n}} \) 4.2. Determine whether \( \left(a_{n}\right) \) is convergent or divergent, and it is convergent, find its limit. 4.3. Determine whether the following series is conditionally convergent absolutely convergent or divergent. 4.4. Find the sum of the series \( \sum_{n=0}^{+\infty} \frac{(-1)^{n} x^{4 n}}{n!} \).

Let's solve each part of the problem step by step. ### 4.1. Find the first 2 terms of the sequence \( (a_n) \) The sequence is defined by the recurrence relation: \[ a_{n+1} = \frac{9^{n+1}}{10^n} \] To find the first two terms, we need to start with an initial term. Let's assume \( a_0 \) is the first term. We can calculate \( a_1 \) and \( a_2 \) as follows: 1. **Calculate \( a_0 \)**: We can choose \( a_0 = 1 \) (or any other value, but 1 is a common choice). 2. **Calculate \( a_1 \)**: \[ a_1 = \frac{9^{1}}{10^{0}} = \frac{9}{1} = 9 \] 3. **Calculate \( a_2 \)**: \[ a_2 = \frac{9^{2}}{10^{1}} = \frac{81}{10} = 8.1 \] Thus, the first two terms of the sequence are: \[ a_0 = 1, \quad a_1 = 9 \] ### 4.2. Determine whether \( (a_n) \) is convergent or divergent, and if convergent, find its limit. To analyze the convergence of the sequence, we can express \( a_n \) in a more explicit form. We can derive a general formula for \( a_n \): Starting from \( a_0 = 1 \): \[ a_1 = 9, \quad a_2 = \frac{9^2}{10^1}, \quad a_3 = \frac{9^3}{10^2}, \ldots \] In general, we can express: \[ a_n = \frac{9^n}{10^{n-1}} = 9 \cdot \left(\frac{9}{10}\right)^{n-1} \] As \( n \to \infty \), the term \( \left(\frac{9}{10}\right)^{n-1} \) approaches 0 because \( \frac{9}{10} < 1 \). Therefore: \[ \lim_{n \to \infty} a_n = 9 \cdot 0 = 0 \] Thus, the sequence \( (a_n) \) is convergent, and its limit is: \[ \lim_{n \to \infty} a_n = 0 \] ### 4.3. Determine whether the following series is conditionally convergent, absolutely convergent, or divergent. The series in question is: \[ \sum_{n=0}^{+\infty} \frac{(-1)^{n} x^{4 n}}{n!} \] To determine the convergence, we can use the ratio test for absolute convergence. Consider the absolute series: \[ \sum_{n=0}^{+\infty} \frac{x^{4n}}{n!} \] Using the ratio test: \[ \frac{a_{n+1}}{a_n} = \frac{x^{4(n+1)}}{(n+1)!} \cdot \frac{n!}{x^{4n}} = \frac{x^4}{n+1} \] Taking the limit as \( n \to \infty \): \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{|x|^4}{n+1} = 0 \] Since this limit is less than 1 for all \( x \), the series converges absolutely for all \( x \). ### 4.4. Find the sum of the series \( \sum_{n=0}^{+\infty} \frac{(-1)^{n} x^{4 n}}{n!} \). This series resembles the Taylor series expansion for the exponential function. Specifically, we can relate it to the series for \( e^x \): \[ e^x = \sum_{n=0}^{+\infty} \frac{x^n}{n!} \] The series we have can be rewritten as: \[ \sum_{n=0}^{+\infty} \frac{(-1)^{n} (x^4)^n}{n!} = e^{-x^4} \] Thus, the sum of the series is: \