Null Sequence Proof: (-1)^n / (n^4 + 3) Explained

Nov 6, 2025 by SLV Team 50 views

$Proving {(-1)^n / (n^4 + 3)} is a Null Sequence$

Hey guys! Today, we're diving into the fascinating world of sequences and proving that a particular sequence is a null sequence. A null sequence, in simple terms, is a sequence that converges to zero. We're going to break down the steps to show that the sequence {(-1)^n / (n^4 + 3)}, where n ranges from 1 to infinity, indeed fits this definition. So, buckle up and let's get started!

Understanding Null Sequences

Before we jump into the specifics, let's make sure we're all on the same page about what a null sequence actually is. In mathematical lingo, a sequence (a_n) is a null sequence if, for any positive number ε (epsilon), no matter how small, there exists a positive integer N such that the absolute value of (a_n) is less than ε for all n greater than N. Phew, that's a mouthful! Basically, it means that as n gets larger and larger, the terms of the sequence get closer and closer to zero. Think of it like a zoomed-in view where the terms of the sequence are hugging zero more and more tightly as you go further down the line. This concept is crucial for many areas of calculus and analysis, so understanding it thoroughly is super important. To put it another way, a sequence converges to zero if its limit as n approaches infinity is zero.

When dealing with sequences, especially those involving fractions and powers, understanding their behavior as n grows infinitely large is key. Does the denominator grow faster than the numerator, causing the fraction to shrink towards zero? Or does some oscillating factor, like (-1)^n, complicate things? These are the questions we need to address. In our case, we have a sequence with an alternating sign due to the (-1)^n term, but the denominator grows much faster than the numerator, suggesting that the sequence will indeed converge to zero. However, a formal proof is required to rigorously establish this convergence. This involves using the epsilon-N definition of a limit, which provides a precise way to show that the terms of the sequence become arbitrarily close to zero as n becomes sufficiently large. So, let's dive into the proof!

The Sequence in Question: {(-1)^n / (n^4 + 3)}

Okay, let's take a closer look at the sequence we're dealing with: {(-1)^n / (n^4 + 3)}. Notice a couple of key things here. First, we have (-1)^n in the numerator. This term alternates between -1 and 1 depending on whether n is odd or even. This means the terms of our sequence will bounce back and forth between positive and negative values. Second, we have n^4 + 3 in the denominator. As n gets larger, this term grows very rapidly. The n^4 part dominates the +3, so the denominator becomes incredibly large as n heads towards infinity. This is a crucial observation, as it suggests that the terms of the sequence will become very small.

But how do we prove that this sequence is a null sequence? We can't just say, "Hey, it looks like it goes to zero!" We need a rigorous mathematical argument. That's where the epsilon-N definition comes in. Remember, we need to show that for any tiny positive number ε, we can find a point in the sequence (an index N) such that all terms after that point are closer to zero than ε. The alternating sign introduced by (-1)^n adds a little twist, but the fast-growing denominator is our main weapon here. Because the denominator grows much faster than the numerator, we suspect that the limit of the sequence as n approaches infinity is indeed zero. This intuition guides our proof, but the formal steps are necessary to turn this intuition into a solid mathematical argument. The challenge lies in finding a suitable N for any given ε, which we will tackle step by step.

The Epsilon-N Proof: Step-by-Step

Alright, let's get down to the nitty-gritty and construct our epsilon-N proof. This type of proof can seem a bit abstract at first, but it's a powerful tool for showing convergence. Here's the plan:

Start with epsilon (ε): We begin by assuming we have an arbitrary positive number ε. Think of ε as a tiny margin of error around zero. We want to show that our sequence eventually falls within this margin and stays there.
Find N: Our goal is to find a positive integer N such that for all n > N, the absolute value of our sequence term |(-1)^n / (n^4 + 3)| is less than ε. This is the heart of the proof – finding a suitable N that works for any given ε.
Show the inequality holds: Once we've found our N, we need to demonstrate that the inequality |(-1)^n / (n^4 + 3)| < ε actually holds for all n > N.

Now, let's put this plan into action. We start with the absolute value of our sequence term: |(-1)^n / (n^4 + 3)|. The absolute value of (-1)^n is always 1, so we can simplify this to 1 / (n^4 + 3). Next, we want to make this expression less than ε. To do this, we can try to find a simple inequality that relates n to ε. The trick is to manipulate the inequality 1 / (n^4 + 3) < ε to isolate n. This process often involves some algebraic manipulation and clever approximations. Once we find a suitable N in terms of ε, we can confidently say that the sequence converges to zero.

Finding the Right N

Okay, so we have |(-1)^n / (n^4 + 3)| = 1 / (n^4 + 3), and we want this to be less than ε. Our mission is to find a value for N that guarantees this inequality holds for all n > N. Here's how we can do it:

Simplify the inequality: We want 1 / (n^4 + 3) < ε. Since n^4 + 3 is always positive, we can take the reciprocal of both sides (and flip the inequality sign) to get n^4 + 3 > 1/ε.
Isolate n^4: Subtract 3 from both sides: n^4 > (1/ε) - 3.
Take the fourth root: Now, take the fourth root of both sides: n > ((1/ε) - 3)^(1/4).

This gives us a clue about what N should be. We need to choose N to be an integer greater than ((1/ε) - 3)^(1/4). However, there's a small catch! What if (1/ε) - 3 is negative? We can avoid this issue by choosing a slightly larger expression that is always positive. Since we are looking for large n, we can safely ignore the -3 term for sufficiently small ε. Thus, a simpler (and often sufficient) choice is to require n > (1/ε)^(1/4). This simplified expression makes it easier to work with and still guarantees that our original inequality holds for large n. The key is to choose N large enough such that the terms of the sequence are within the desired margin of error ε. So, let's formalize this.

Formalizing the Choice of N

So, we've figured out that we need n > (1/ε)^(1/4). This means we can choose N to be any integer greater than (1/ε)^(1/4). To be absolutely precise, we can define N as the smallest integer greater than (1/ε)^(1/4). This is often written using the ceiling function: N = ⌈(1/ε)^(1/4)⌉. The ceiling function simply rounds a number up to the nearest integer. This ensures that N is always an integer and satisfies our condition.

Now, let's recap. Given any ε > 0, we've chosen N = ⌈(1/ε)^(1/4)⌉. This means that for any n > N, we have n > (1/ε)^(1/4). Raising both sides to the fourth power gives us n^4 > 1/ε. Adding 3 to both sides gives n^4 + 3 > (1/ε). Taking the reciprocal of both sides (and flipping the inequality) gives 1 / (n^4 + 3) < ε. Since |(-1)^n / (n^4 + 3)| = 1 / (n^4 + 3), we have shown that |(-1)^n / (n^4 + 3)| < ε for all n > N. This is exactly what we needed to show! We've successfully found an N that works for any given ε, which is the essence of the epsilon-N definition of a limit.

Showing the Inequality Holds

Now that we've chosen our N, we need to formally show that the inequality |(-1)^n / (n^4 + 3)| < ε holds for all n > N. This step is crucial to complete our proof and demonstrate that our choice of N was indeed correct. Let's walk through the logic step by step.

We start with our chosen N = ⌈(1/ε)^(1/4)⌉. Since n > N, we know that n > (1/ε)^(1/4). Raising both sides to the fourth power, we get n^4 > 1/ε. Now, let's add 3 to both sides: n^4 + 3 > (1/ε). Taking the reciprocal of both sides (and flipping the inequality sign) gives us 1 / (n^4 + 3) < ε. But remember that |(-1)^n / (n^4 + 3)| is simply equal to 1 / (n^4 + 3). So, we have shown that |(-1)^n / (n^4 + 3)| < ε for all n > N. This is precisely what we wanted to prove!

We've successfully demonstrated that for any arbitrarily small positive number ε, we can find a positive integer N such that the absolute value of the terms of our sequence is less than ε for all n greater than N. This rigorously proves that the sequence {(-1)^n / (n^4 + 3)} is indeed a null sequence. The key takeaway here is the systematic approach of the epsilon-N proof: starting with ε, finding a suitable N, and then verifying that the inequality holds. This technique is fundamental in real analysis and provides a powerful way to establish convergence rigorously.

Conclusion

So, there you have it! We've successfully proven that the sequence {(-1)^n / (n^4 + 3)} is a null sequence using the epsilon-N definition. This might seem a bit abstract, but it's a fundamental concept in mathematics. Remember, the key is that as n gets really big, the terms of the sequence get closer and closer to zero. Hopefully, this step-by-step explanation has made the process a bit clearer. Keep practicing these types of proofs, and you'll become a sequence-busting pro in no time! Understanding null sequences is a stepping stone to more advanced topics in calculus and analysis, so the effort you put in now will definitely pay off in the long run. Great job, guys!