Question

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit. $$ a_{n}=\frac{2^{n+1}}{5^{n}} $$

Short answer

The sequence is bounded, monotonic decreasing, and convergent with a limit of 0.

Step-by-step solution

Simplify the Sequence

The given sequence is \( a_n = \frac{2^{n+1}}{5^n} \). Simplify this to \( a_n = \frac{2 \cdot 2^n}{5^n} = 2 \left( \frac{2}{5} \right)^n \).

Check Monotonicity

To check monotonicity, consider \( \left( \frac{2}{5} \right) \), which is less than 1. Observe that as \( n \) increases, \( \left( \frac{2}{5} \right)^n \) will decrease because raising a fraction less than 1 to higher powers gets smaller. Therefore, the sequence is monotonic decreasing.

Determine Boundedness

Since \( \left( \frac{2}{5} \right)^n \) decreases to 0 as \( n \to \infty \), and is always non-negative, the sequence is bounded below by 0. Additionally, since it's decreasing, its maximum is its first term which is 4.

Test for Convergence

A sequence is convergent if it approaches a finite limit as \( n \to \infty \). Since \( \left( \frac{2}{5} \right)^n \to 0 \), we have \( a_n = 2 \left( \frac{2}{5} \right)^n \to 0 \). Thus, the sequence is convergent.

Find the Limit

As previously determined, the limit of \( a_n = 2 \left( \frac{2}{5} \right)^n \) as \( n \to \infty \) is 0. Therefore, the sequence converges to 0.

Key concepts

Bounded Sequences

A sequence is said to be **bounded** if it is confined within a certain range of values, particularly having an upper and lower limit. This means that there exists some numbers, say \( M \) and \( m \), such that every element of the sequence is less than or equal to \( M \) and greater than or equal to \( m \).
For the sequence \( a_n = 2 \left( \frac{2}{5} \right)^n \), let us examine its bounds. This sequence has a lower bound of 0, as each term always remains non-negative since fractions to positive integer powers do not change sign.
Because the sequence is decreasing, the upper bound is simply the first term when \( n = 0 \), which calculates to 4. Thus, this sequence is bounded with bounds 0 and 4.
Ultimately, recognizing this can help greatly in further analyzing the sequence's characteristics.

Monotonic Sequences

A **monotonic sequence** is one that consistently increases or decreases as its index increases. In mathematical terms, a sequence \( \{a_n\} \) is monotonic if either \( a_{n+1} \geq a_n \) for all \( n \) (monotonic increasing) or \( a_{n+1} \leq a_n \) for all \( n \) (monotonic decreasing).
The sequence \( a_n = 2 \left( \frac{2}{5} \right)^n \) is shown to be **monotonic decreasing**. This is because \( \left( \frac{2}{5} \right) < 1 \), making each subsequent term smaller than the previous one as \( n \) increases.

• When \( n =0 \), \( a_0 = 2 \).
• As \( n \to \infty \), the terms shrink towards 0.

Being able to identify a sequence as monotonic helps us predict its behavior as well, often aiding in determining convergence or divergence.

Limit of a Sequence

The **limit of a sequence** is the value that the terms of the sequence approach as the index \( n \) gets infinitely large. A sequence is convergent if it has a limit, and divergent if it does not.
In our sequence, \( a_n = 2 \left( \frac{2}{5} \right)^n \), we calculate its limit as \( n \to \infty \). As shown in the analysis, since \( \left( \frac{2}{5} \right)^n \) approaches 0, and multiplying by 2 does not change the fact that it approaches 0, the entire sequence tends to 0.
We now understand that this sequence converges to 0. Recognizing the limit of a sequence is crucial, as it tells us about the long-term behavior of the sequence and whether it stabilizes around a particular value.