Sequence $v_n$: Proofs, Interpretations, And Discussion

$$

Hey guys! Let's dive into the fascinating world of sequences with a detailed look at the sequence v_n = rac{2n+1}{n+1}. We're going to break down this sequence, prove some key properties, interpret the results, and have a good discussion about it. So, buckle up and let's get started!

a) Proving $v_n < 2$ and Interpreting the Result

So, the first thing we need to show is that every term in the sequence $v_n$ is less than 2. How do we do that? Well, let’s start by looking at the inequality we want to prove:

$v_n < 2$

Now, substitute the formula for $v_n$:

rac{2n+1}{n+1} < 2

To get rid of the fraction, we can multiply both sides by $(n+1)$. Since $n$ is a natural number, $(n+1)$ is always positive, so we don’t need to worry about flipping the inequality sign:

$2n+1 < 2(n+1)$

Expand the right side:

$2n+1 < 2n+2$

Now, subtract $2n$ from both sides:

$1 < 2$

Wow, look at that! We ended up with a statement that is always true. This means our original inequality, $v_n < 2$, is also true for all natural numbers $n$. Isn't that neat?

Interpreting the Result

Okay, so we’ve proved that $v_n$ is always less than 2. But what does this actually mean? In simple terms, it tells us that the sequence $v_n$ is bounded above by 2. Think of it like this: no matter how far along we go in the sequence, the terms will never reach or exceed the value 2. They'll get closer and closer, but they'll never quite get there.

This is a crucial piece of information because it gives us an idea of the sequence's behavior. It's like knowing the speed limit on a road – we know the cars can't go faster than that. Similarly, knowing that $v_n < 2$ helps us understand the upper limit of our sequence.

In the context of sequences, having an upper bound is super important. It's a key factor when we start talking about convergence, which is basically whether a sequence approaches a specific value as $n$ gets really, really big. This little proof is a foundational step in understanding the bigger picture of this sequence.

b) Proving the Sequence is Bounded Below by 1

Alright, so we know our sequence $v_n$ is like a race car that can't exceed a speed of 2. But what about the lower limit? Is there a minimum value that the terms of the sequence will always be greater than or equal to? Let's find out by proving that the sequence is bounded below by 1.

To prove this, we need to show that:

$v_n ext{greater than or equal to} 1$

Again, let’s substitute the formula for $v_n$:

rac{2n+1}{n+1} ext{greater than or equal to} 1

Multiply both sides by $(n+1)$ (remember, this is positive, so no flipping):

$2n+1 ext{greater than or equal to} n+1$

Now, let's isolate $n$ by subtracting $n$ from both sides:

$n+1 ext{greater than or equal to} 1$

And finally, subtract 1 from both sides:

$n ext{greater than or equal to} 0$

This statement is true for all natural numbers $n$ (since natural numbers start at 0 or 1, depending on the convention). This confirms that our inequality, $v_n ext{greater than or equal to} 1$, holds true for all $n ext{ in } ext{N}$.

What Does It Mean to Be Bounded Below?

So, we've shown that our sequence is like a car that can't go slower than a certain speed. Being bounded below by 1 means that the terms of the sequence will always be at least 1. They might get bigger, but they'll never dip below that 1 mark. This gives us a sense of the sequence's lower limit, just like the upper bound gave us the higher limit.

Having both an upper and a lower bound is fantastic news for a sequence. It means the sequence is bounded. Bounded sequences are much easier to work with and analyze. They're like having guardrails on a road – we know the sequence won't go completely wild and head off to infinity (or negative infinity!).

Knowing that $v_n$ is bounded between 1 and 2 helps us visualize its behavior even more clearly. It’s like having a clear range within which our sequence operates. This is super helpful as we move towards discussing the sequence's convergence.

c) Discussing the Sequence

Now that we've established some key properties of our sequence $v_n$, let's dive into a discussion about its overall behavior. We know that it's bounded – it's trapped between 1 and 2. But what else can we say about it? Is it increasing? Is it decreasing? Does it approach a specific value as $n$ gets larger?

Monotonicity: Is the Sequence Increasing or Decreasing?

The first thing we might want to investigate is whether the sequence is monotonic. Monotonic sequences are either always increasing or always decreasing. To figure this out, we can look at the difference between consecutive terms: $v_{n+1} - v_n$.

Let's calculate $v_{n+1}$:

v_{n+1} = rac{2(n+1)+1}{(n+1)+1} = rac{2n+3}{n+2}

Now, let's find the difference $v_{n+1} - v_n$:

v_{n+1} - v_n = rac{2n+3}{n+2} - rac{2n+1}{n+1}

To subtract these fractions, we need a common denominator, which is $(n+2)(n+1)$. So, we rewrite the fractions:

v_{n+1} - v_n = rac{(2n+3)(n+1)}{(n+2)(n+1)} - rac{(2n+1)(n+2)}{(n+2)(n+1)}

Now, expand the numerators:

v_{n+1} - v_n = rac{2n^2 + 5n + 3}{(n+2)(n+1)} - rac{2n^2 + 5n + 2}{(n+2)(n+1)}

Combine the fractions:

v_{n+1} - v_n = rac{(2n^2 + 5n + 3) - (2n^2 + 5n + 2)}{(n+2)(n+1)}

Simplify:

v_{n+1} - v_n = rac{1}{(n+2)(n+1)}

Since $n$ is a natural number, $(n+2)(n+1)$ is always positive. Therefore, $v_{n+1} - v_n$ is always positive. This means that $v_{n+1} > v_n$ for all $n$. So, our sequence is increasing!

Convergence: What Happens as n Gets Really Big?

We know our sequence is bounded above by 2, bounded below by 1, and it's increasing. What does this tell us about its long-term behavior? Well, it strongly suggests that the sequence converges. Convergence means that the terms of the sequence get closer and closer to a specific value as $n$ approaches infinity.

To find the limit, we can look at what happens to $v_n$ as $n$ gets extremely large. We can rewrite $v_n$ like this:

v_n = rac{2n+1}{n+1} = rac{n(2 + rac{1}{n})}{n(1 + rac{1}{n})}

Now, cancel out the $n$:

v_n = rac{2 + rac{1}{n}}{1 + rac{1}{n}}

As $n$ approaches infinity, rac{1}{n} approaches 0. So, we have:

$ ext{lim (as n approaches infinity)} v_n = rac{2 + 0}{1 + 0} = 2$

Aha! The sequence converges to 2. This makes perfect sense given what we've already discovered: the sequence is increasing and bounded above by 2, so it's naturally going to creep closer and closer to 2 without ever quite reaching it.

Putting It All Together: A Holistic View of the Sequence

Let's recap what we've learned about the sequence v_n = rac{2n+1}{n+1}:

• It is bounded above by 2.
• It is bounded below by 1.
• It is increasing (monotonic).
• It converges to 2.

By proving these properties and interpreting them, we've built a comprehensive understanding of this sequence. We know its limits, its direction, and its long-term behavior. This kind of analysis is super important in many areas of mathematics and science, from calculus to physics to computer science.

So, there you have it, guys! We've taken a deep dive into this sequence, and hopefully, you've gained some valuable insights into how to analyze and understand sequences in general. Keep exploring, keep questioning, and keep learning!