Question

Qno.2) Establish either the sequence (X=Xn ) converges or diverges where
I) Xn=n/n+1 ii) Xn= (_1^n)n/n+1 iii)Xn= n^2/n+1
B) find the limits of the sequences
I) (2+1/n)^2 ii) (-1)^n/ n+2. iii) (n)^1/2-1/(n)^1/2+1 iv)n+1/n(n)^1/2
V)a^n+1+b^n+1/a^n+b^n for 0<a<b .

Answer 1

Answer on Question #82486 – Math – Real Analysis

1. Establish either the sequence $X = x_{n}$ converges or diverges where

Question

i)

$$x _ {n} = \frac {n}{n + 1}$$

Solution

$$\lim _ {n \rightarrow \infty} x _ {n} = \lim _ {n \rightarrow \infty} \frac {n}{n + 1} = 1$$

**Answer:** the sequence converges

Question

ii)

$$x _ {n} = \frac {(- 1) ^ {n} n}{n + 1}$$

Solution

$$\lim _ {n \rightarrow \infty} x _ {n} = \lim _ {n \rightarrow \infty} \frac {n}{n + 1} = 1 \text{ for even } n$$ $$\lim _ {n \rightarrow \infty} x _ {n} = \lim _ {n \rightarrow \infty} \frac {- n}{n + 1} = - 1 \text{ for odd } n$$

So, the sequence has not limit.

**Answer:** the sequence diverges

Question

iii)

$$x _ {n} = \frac {n ^ {2}}{n + 1}$$

Solution

$$\lim _ {n \to \infty} x _ {n} = \lim _ {n \to \infty} \frac {n ^ {2}}{n + 1} = \lim _ {n \to \infty} \left(n - 1 + \frac {1}{n + 1}\right) = \infty$$

Answer: the sequence diverges

2. Find the limits of the sequences

Question

i)

$$\left(2 + \frac {1}{n}\right) ^ {2}$$

Solution

$$\lim _ {n \to \infty} \left(2 + \frac {1}{n}\right) ^ {2} = 2 ^ {2} = 4$$

Answer: 4.

Question

ii)

$$\frac {(- 1) ^ {n}}{n + 2}$$

Solution

$$\lim _ {n \to \infty} \frac {(- 1) ^ {n}}{n + 2} = 0$$

Answer: 0.

Question

iii)

$$\frac {n ^ {1 / 2} - 1}{n ^ {1 / 2} + 1}$$

Solution

$$\lim _ {n \to \infty} \frac {n ^ {1 / 2} - 1}{n ^ {1 / 2} + 1} = \lim _ {n \to \infty} \frac {n ^ {1 / 2} (1 - 1 / n ^ {1 / 2})}{n ^ {1 / 2} (1 + 1 / n ^ {1 / 2})} = \frac {1 - 0}{1 + 0} = 1.$$

Answer: 1.

iv)

Question

$$\frac {n + 1}{n n ^ {1 / 2}}$$

Solution

$$\lim _ {n \to \infty} \frac {n + 1}{n n ^ {1 / 2}} = \lim _ {n \to \infty} \frac {1 + 1 / n}{n ^ {1 / 2}} = 0$$

Answer: 0.

Question

v)

$$\frac {a ^ {n + 1} + b ^ {n + 1}}{a ^ {n} + b ^ {n}} \text{ for } 0 < a < b$$

Solution

\lim _ {n \to \infty} \frac {a ^ {n + 1} + b ^ {n + 1}}{a ^ {n} + b ^ {n}} = \lim _ {n \to \infty} \frac {\frac {a ^ {n + 1}}{\frac {b ^ {n + 1} + b ^ {n + 1}}{a ^ {n}}} - \frac {1}{\frac {1}{b}} = b.

Answer: b.

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