Question

Определите какой из рядов сходится и указать тип сходимости. 1) (-1)^(n+1)*2^n /3^n+n 2) (-1)^n*n^4/n^4+3n^4+1 Помогите, пожалуйста.

Answer 1

$\sum \limits _{n=1}^{\infty }\frac{(-1)^{n+1}\cdot 2^{n}}{3^{n}+n}\\\\Priznak\; Lejbnica:\; \; a)\; \lim\limits _{n \to \infty} |a_n |= \lim\limits _{n \to \infty}\, \frac{2^{n}}{3^{n}+n}= \lim\limits _{n \to \infty}\, \frac{2^{n}\cdot ln2}{3^{n}\cdot ln3+1}=\\\\=\lim\limits _{n \to \infty}\,\frac{2^{n}\cdot ln^22}{3^{n}\cdot ln^23}=\frac{ln^22}{ln^23}\cdot \lim\limits _{n \to +\infty}\, \Big (\frac{2}{3}\Big )^{n}=\frac{ln^22}{ln^23}\cdot 0=0$

$b)\; \; |a_1|\ \textgreater \ |a_2|\ \textgreater \ a_3|\ \textgreater \ \ldots \ \textgreater \ |a_{n}|\ \textgreater \ \ldots \\\\\frac{2}{3+1}\ \textgreater \ \frac{2^2}{3^2+1}\ \textgreater \ \frac{2^3}{3^3+1}\ \textgreater \ \ldots \ \textgreater \ \frac{2^{n}}{3^{n}+1}\ \textgreater \ \ldots$

Знакочередующийся ряд сходится по признаку Лейбница.
Ряд будет абсолютно сходящимся, т.к. для ряда, составленного из абсолютных величин, выполняется признак Даламбера:


$\lim\limits _{n \to \infty}\frac{|a_{n+1}|}{|a_{n}|}= \lim\limits _{n \to \infty}(\frac{2^{n+1}}{3^{n+1}+n+1}:\frac{2^{n}}{3^{n}+1})= \lim\limits _{n \to \infty}\frac{2^{n}\cdot 2(3^{n}+1)}{(3^{n+1}+n+1)\cdot 2^{n}}= \\\\=\lim\limits _{n \to \infty}\frac{2(3^{n}+1)}{3^{n+1}+n+1}=\lim\limits _{n \to \infty}\frac{2\cdot 3^{n}\cdo ln3}{3^{n+1}\cdot ln3+1}=\lim\limits _{n \to \infty}\frac{2\cdot 3^{n}\cdot ln^23}{3^{n+1}\cdot ln^23}=\\\\=\lim\limits _{n \to \infty}\frac{2\cdot 3^{n}ln^23}{3^{n}\cdot 3ln^23}=\frac{2}{3}<1$


$2)\; \; \sum \limits _{n=1}^{\infty }\frac{(-1)^{n}\cdot n^4}{n^4+3n^2+1}\\\\Priznak\; Lejbnica:\\\\a)\; \lim\limits _{n\to \infty }|a_{n}|=\lim\limits _{n \to \infty}\frac{n^4}{n^4+3n^2+1}=\lim\limits _{n \to \infty}\, \frac{1}{1+\frac{3}{n^2}+\frac{1}{n^4}}=1\ne 0\; \; \Rightarrow$

так как не выполняется 1 пункт признака Лейбница, то нет смысла проверять 2 пункт. Вывод: знакочередующийся ряд расходится.

Answer 2

Task: Determine the convergence and specify the type of convergence for the given series.

To determine the convergence and type of convergence for the given series, let's analyze each series separately.

Series 1: The first series is given by: (-1)^(n+1) * 2^n / (3^n + n + 2)

To determine the convergence of this series, we can use the ratio test. The ratio test states that if the absolute value of the ratio of consecutive terms approaches a limit less than 1 as n approaches infinity, then the series converges. If the limit is greater than 1, the series diverges, and if the limit is equal to 1, the test is inconclusive.

Let's apply the ratio test to the first series: lim(n->∞) |((-1)^(n+2) * 2^(n+1) / (3^(n+1) + (n+1) + 2)) / ((-1)^(n+1) * 2^n / (3^n + n + 2))|

Simplifying the expression, we get: lim(n->∞) |(-2 * 2^n * (3^n + n + 2)) / (2^n * (3^(n+1) + (n+1) + 2))|

Canceling out the common terms, we have: lim(n->∞) |-2 / (3 * (1 + (n+1)/3^n + 2/3^n))|

As n approaches infinity, the terms (n+1)/3^n and 2/3^n both approach 0, so the limit simplifies to: |-2 / (3 * (1 + 0 + 0))|

The absolute value of the limit is 2/3, which is less than 1. Therefore, by the ratio test, the first series converges.

Now let's determine the type of convergence. To do this, we need to analyze the behavior of the terms in the series. The terms in the series alternate in sign due to the (-1)^(n+1) factor. Additionally, the terms decrease in magnitude as n increases. This indicates that the series has an alternating and decreasing behavior.

Therefore, the first series converges by the ratio test and has alternating and decreasing convergence.

Series 2: The second series is given by: (-1)^n * n^4 / (n^4 + 3n^4 + 1)

To determine the convergence of this series, we can again use the ratio test. Let's apply the ratio test to the second series: lim(n->∞) |((-1)^(n+1) * (n+1)^4 / ((n+1)^4 + 3(n+1)^4 + 1)) / ((-1)^n * n^4 / (n^4 + 3n^4 + 1))|

Simplifying the expression, we get: lim(n->∞) |(-1 * (n+1)^4 * (n^4 + 3n^4 + 1)) / (n^4 * ((n+1)^4 + 3(n+1)^4 + 1))|

Canceling out the common terms, we have: lim(n->∞) |-1 / ((n+1)^4 + 3(n+1)^4 + 1)|

As n approaches infinity, the terms (n+1)^4 and 3(n+1)^4 dominate the expression, so the limit simplifies to: |-1 / (4(n+1)^4)|

The absolute value of the limit is 1/4, which is less than 1. Therefore, by the ratio test, the second series converges.

Now let's determine the type of convergence. Similar to the first series, the terms in the second series alternate in sign due to the (-1)^n factor. Additionally, the terms decrease in magnitude as n increases. This indicates that the series has an alternating and decreasing behavior.

Therefore, the second series converges by the ratio test and has alternating and decreasing convergence.

To summarize: - Series 1 converges with alternating and decreasing convergence. - Series 2 converges with alternating and decreasing convergence.

Please let me know if there's anything else I can help you with!