The sequence \(((1+\frac{1}{n})^\frac{1}{n})\) is1. bounded2. unboun

1.	The sequence \(((1+\frac{1}{n})^\frac{1}{n})\) is1. bounded2. unbounded3. only lower bounded4. only upper bounded
Answer» Correct Answer - Option 1 : bounded Concept: A sequence {a_n} is said to be bounded below if there exists a number l such that l ≤ a_n ∀ n ≤ N. Such a number l is called a lower bound of {an}. A sequence {an} s said to be bounded above if there exists a number u such that a_n ≤ u, ∀ n ≤ N such a number u is called an upper bound of {an}. When a sequence {a_n}, n∈N a is both bounded below and bounded above, it is called a bounded sequence. Calculation: First, let us check the boundedness of the sequence from below. Note that ∀n∈N \((1+\frac{1}{n})>1\) ⇒ \((1+\frac{1}{n})^\frac{1}{n}>1\) Hence the sequence is bounded below. Now we check the boundedness of the sequence from above. You can see that for all n ∈ N \(\frac{1}{n}≤1\) ⇒ \(1+\frac{1}{n}≤2≤2^n \) ⇒ \((1+\frac{1}{n})^\frac{1}{n}≤2\) Thus the sequence is bounded above also. Hence the sequence is bounded.

The sequence \(((1+\frac{1}{n})^\frac{1}{n})\) is1. bounded2. unbounded3. only lower bounded4. only upper bounded

Answer» Correct Answer - Option 1 : bounded

Concept:

A sequence {a_n} is said to be bounded below if there exists a number l such that l ≤ a_n ∀ n ≤ N. Such a number l is called a lower bound of {an}.

A sequence {an} s said to be bounded above if there exists a number u such that a_n ≤ u, ∀ n ≤ N such a number u is called an upper bound of {an}.
When a sequence {a_n}, n∈N a is both bounded below and bounded above, it is called a bounded sequence.

Calculation:

First, let us check the boundedness of the sequence from below. Note that ∀n∈N

\((1+\frac{1}{n})>1\) ⇒ \((1+\frac{1}{n})^\frac{1}{n}>1\)

Hence the sequence is bounded below. Now we check the boundedness of the sequence from above. You can see that for all n ∈ N

\(\frac{1}{n}≤1\)

⇒ \(1+\frac{1}{n}≤2≤2^n \)

⇒ \((1+\frac{1}{n})^\frac{1}{n}≤2\)

Thus the sequence is bounded above also. Hence the sequence is bounded.