Question

Q. For xn given by the following formulas, establish either the convergence or divergence of the sequence X=(xn)
(a) xn= n/(n+1), (b) xn=(〖(-1)〗^n n)/(n+1) ,
(c) xn=n2/(n+1), (d)xn=(2n2+3)/(n2+1)

Answer 1

Answer on Question #64521 – Math – Real Analysis

Question

For $x_n$ given by the following formulas, establish either the convergence or divergence of the sequence

(a) $x_n = \frac{n}{n+1}$,

(b) $x_n = \frac{(-1)^n n}{n+1}$,

(c) $x_n = \frac{n^2}{n+1}$,

(d) $x_n = \frac{2n^2 + 3}{n^2 + 1}$

Solution (a). Let's consider the sequence $x_n = \frac{n}{n+1}$, $n \geq 1$. Its limit is

$$\lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{n}{n} \cdot \frac{1}{1 + \frac{1}{n}} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} = 1$$

Thus, the sequence $x_n = \frac{n}{n+1}$ is convergent.

Answer: (a) $x_n = \frac{n}{n+1}$ is convergent.

Solution (b). Let's consider the subsequences $x_{2k} = \frac{(-1)^{2k} 2k}{2k + 1} = \frac{2k}{2k + 1} = 1 - \frac{1}{2k + 1}$ and $x_{2k-1} = \frac{(-1)^{2k-1} (2k-1)}{2k-1 + 1} = \frac{1 - 2k}{2k} = \frac{1}{2k} - 1$ of the sequence $x_n = \frac{(-1)^n n}{n+1}$.

Since $\lim_{k\to \infty}x_{2k} = \lim_{k\to \infty}\left(1 - \frac{1}{2k + 1}\right) = 1$ and $\lim_{k\to \infty}x_{2k - 1} = \lim_{k\to \infty}\left(\frac{1}{2k} -1\right) = -1$, then the sequence $x_{n} = \frac{(-1)^{n}n}{n + 1}$ has two different limit points. Therefore $x_{n} = \frac{(-1)^{n}n}{n + 1}$ is divergent.

Answer: (b) $x_n = \frac{(-1)^n n}{n + 1}$ is divergent.

Solution (c). Since $x_n = \frac{n^2}{n+1} = \frac{n^2 - 1 + 1}{n+1} = n - 1 + \frac{1}{n+1} > n - 1$ for all natural $n$ and sequence $y_n = n - 1$ is unbounded from above. Hence $x_n = \frac{n^2}{n+1}$ is unbounded, $x_n = \frac{n^2}{n+1}$ is divergent.

Answer: (c) $x_n = \frac{n^2}{n+1}$ is divergent.

Solution (d). Let's consider the sequence $x_n = \frac{2n^2 + 3}{n^2 + 1}$, $n \geq 1$. Its limit is

$$\lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{2n^2 + 3}{n^2 + 1} = \lim_{n \to \infty} \frac{n^2}{n^2} \cdot \frac{2 + \frac{3}{n^2}}{1 + \frac{1}{n}} = \lim_{n \to \infty} \frac{2 + \frac{3}{n^2}}{1 + \frac{1}{n}} = 2.$$

So, the sequence $x_n = \frac{2n^2 + 3}{n^2 + 1}$ is convergent.

Answer: (d) $x_n = \frac{2n^2 + 3}{n^2 + 1}$ is convergent.

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