Show whether the sequence converges or diverges, is monotonic, and bounded. If it is bounded, give the greatest lower bound and least upper bound in the form of an inequality.

Understand the Problem

The question asks to analyze three sequences to determine whether they converge or diverge, whether they are monotonic, and whether they are bounded. If bounded, the question requires finding the greatest lower bound and least upper bound in inequality form.

Answer

1. Diverges, monotonic increasing, not bounded. 2. Diverges, not monotonic, bounded by $ -1 < a_n < 0 $. 3. Converges to $ \frac{1}{2} $, monotonic increasing, bounded by $ 0 < a_n < \frac{1}{2} $.

Steps to Solve

Sequence Analysis of ( a_n = \frac{3^n}{n} )

Examine the behavior of the sequence as ( n ) approaches infinity.

The exponential function ( 3^n ) grows faster than the linear function ( n ).
Thus, we can conclude:

$$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{3^n}{n} = \infty $$

This indicates the sequence diverges.

Monotonicity of ( a_n = \frac{3^n}{n} )

To check if the sequence is increasing or decreasing, consider the ratio:

$$ \frac{a_{n+1}}{a_n} = \frac{3^{n+1}/(n+1)}{3^n/n} = \frac{3n}{n+1} $$

As ( n \to \infty ), this approaches ( 3 ), meaning:

For large ( n ), ( a_{n+1} > a_n ).
Therefore, the sequence is increasing (monotonic).

Boundedness of ( a_n = \frac{3^n}{n} )

Since the sequence diverges (goes to infinity), it cannot be bounded. Thus:

Not bounded.

Sequence Analysis of ( a_n = \frac{(-1)^n n}{n+1} )

Determine the limit as ( n ) approaches infinity.

Since the sequence oscillates based on the sign:

$$ a_n \to \begin{cases} 0 & \text{if } n \text{ is even} \ -1 & \text{if } n \text{ is odd} \end{cases} $$

Thus, the sequence does not converge.

Monotonicity of ( a_n = \frac{(-1)^n n}{n+1} )

The sequence alternates between positive and negative values:

Thus, it is neither increasing nor decreasing (not monotonic).

Boundedness of ( a_n = \frac{(-1)^n n}{n+1} )

The values oscillate between ( 0 ) (as the limiting case) and approximately ( -1 ):

The sequence is bounded between ( -1 ) and ( 0 ).

$$ -1 < a_n < 0 $$

Sequence Analysis of ( a_n = \frac{n}{2n+1} )

Determine the limit as ( n ) approaches infinity:

$$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n}{2n+1} = \lim_{n \to \infty} \frac{1}{2 + \frac{1}{n}} = \frac{1}{2} $$

Thus, the sequence converges to ( \frac{1}{2} ).

Monotonicity of ( a_n = \frac{n}{2n+1} )

Check the first derivative or the ratio:

$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)/(2(n+1)+1)}{n/(2n+1)} $$

As ( n ) increases, this suggests ( a_n ) is increasing, making the sequence monotonic.

Boundedness of ( a_n = \frac{n}{2n+1} )

The sequence approaches ( \frac{1}{2} ), thus it is bounded:

$$ 0 < a_n < \frac{1}{2} $$

For ( a_n = \frac{3^n}{n} ): Diverges, monotonic increasing, not bounded.
For ( a_n = \frac{(-1)^n n}{n+1} ): Diverges, not monotonic, bounded by ( -1 < a_n < 0 ).
For ( a_n = \frac{n}{2n+1} ): Converges to ( \frac{1}{2} ), monotonic increasing, bounded by ( 0 < a_n < \frac{1}{2} ).

More Information

The sequences exhibit different behaviors. The first diverges; the second oscillates but remains bounded; the third converges and is bounded as well. Understanding the growth of functions, oscillatory behavior, and limit processes is crucial in sequence analysis.

Tips

Assuming that a sequence is bounded if it converges; all converging sequences are bounded, but not all bounded sequences converge (like in the second case).
Ignoring the behavior of sequences at infinity can lead to misidentifying convergence.

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