Finding N For An > C And Limit Proof

Hey guys! Today, we're diving into some cool math problems related to sequences and limits. Specifically, we'll tackle the challenge of finding a value N such that for all n greater than N, a given sequence a_n exceeds a certain value C. Plus, we'll prove a fundamental limit theorem. Let's break it down step by step.

3.9. Determining N for $a_n > C$

So, the main goal here is to figure out, given a number C that's bigger than zero, what value of N will guarantee that every term a_n in our sequence, where n is greater than N, is also greater than C. We're given three different formulas for a_n, so let's look at each one individually.

a) $a_n = lg^2 n$

Let's start with the logarithmic sequence $a_n = lg^2 n$. Here, lg usually represents the base-10 logarithm (log base 10). We need to find N such that for all n > N, $lg^2 n > C$. In other words, we want to find out when the square of the base-10 logarithm of n is greater than C. The key idea is to isolate n. First, take the square root of both sides of the inequality, so we get $|lg n| >

\sqrt{C}$. Since lg n can be positive or negative, we consider two cases. If lg n is positive, then *lg n >

\sqrt{C}*, which means n > 10^{\sqrt{C}}. If lg n is negative, then *lg n < -

\sqrt{C}*, which means n < 10^{-\sqrt{C}}. However, since we are looking for n > N, we only consider the first case n > 10^{\sqrt{C}}}. Therefore, we can choose N = 10^{\sqrt{C}}}. So, for any n > N = 10^{\sqrt{C}}}, the condition $a_n > C$ will be true. To summarize, if we pick N to be $10^{\sqrt{C}}$, then for any n greater than N, $lg^2 n$ will be bigger than C. This makes sense because as n grows, its logarithm grows, and squaring it makes it grow even faster!

b) $a_n =

\sqrt{n^4-4n2+5}$

Next up, we've got the sequence with a square root: $a_n =

\sqrt{n^4-4n2+5}$. We want to find N such that for all n > N, $

\sqrt{n^4-4n2+5} > C$. First, let's square both sides of the inequality to get rid of the square root: $n^4 - 4n^2 + 5 > C^2$. Now, rearrange the inequality: $n^4 - 4n^2 + (5 - C^2) > 0$. Let's make a substitution to simplify this. Let $x = n^2$. Then we have $x^2 - 4x + (5 - C^2) > 0$. This is a quadratic inequality. To find when this inequality holds, we can first find the roots of the corresponding quadratic equation $x^2 - 4x + (5 - C^2) = 0$. Using the quadratic formula, we get:

$x = \frac{-(-4) \pm

\sqrt{(-4)^2 - 4(1)(5 - C^2)}}{2(1)} = \frac{4 \pm

\sqrt{16 - 20 + 4C^2}}{2} = \frac{4 \pm

\sqrt{4C^2 - 4}}{2} = 2 \pm

\sqrt{C^2 - 1}$

So, $x = 2 +

\sqrt{C^2 - 1}$ or $x = 2 -

\sqrt{C^2 - 1}$. Since $x = n^2$, we have $n^2 = 2 +

\sqrt{C^2 - 1}$ or $n^2 = 2 -

\sqrt{C^2 - 1}$. Thus, $n =

\sqrt{2 +

\sqrt{C^2 - 1}}$ or $n =

\sqrt{2 -

\sqrt{C^2 - 1}}$. Since we want $n^4 - 4n^2 + (5 - C^2) > 0$, n must be greater than $

\sqrt{2 +

\sqrt{C^2 - 1}}$. Therefore, we can choose $N =

\sqrt{2 +

\sqrt{C^2 - 1}}$. Note that this is only valid when $C > 1$. If $C

\leq 1$, we need to consider the original inequality more carefully. However, for $C > 1$, if we pick N to be $

\sqrt{2 +

\sqrt{C^2 - 1}}$, then for any n greater than N, $

\sqrt{n^4-4n2+5}$ will be bigger than C. This one is a bit trickier, but breaking it down into smaller steps makes it manageable.

c) $a_n = \frac{n^2}{n+sin n}$

Finally, let's tackle the rational sequence: $a_n = \frac{n^2}{n+sin n}$. We need to find N such that for all n > N, $\frac{n^2}{n+sin n} > C$. We can rewrite the inequality as $n^2 > C(n + sin n)$, or $n^2 > Cn + C sin n$. Since $-1

\leq sin n

\leq 1$, we have $-C

\leq C sin n

\leq C$. Therefore, $Cn + C sin n > Cn - C$. So, if we can ensure that $n^2 > Cn$, then we are getting closer to $a_n > C$. This would mean that $n > C$. Now, let's examine $a_n = \frac{n^2}{n + sin n} = \frac{n}{1 + \frac{sin n}{n}}$. As n approaches infinity, $\frac{sin n}{n}$ approaches 0, so a_n behaves like n for large n. Therefore, for large n, $a_n \approx n$. Thus, we want to find N such that for all n > N, n > C. This suggests that we can choose N = C. However, we must be more rigorous. Let's write $n^2 > C(n + sin n)$ as $n^2 - Cn > C sin n$. Since $|C sin n|

\leq |C|$, if $n^2 - Cn > |C|$, then the inequality holds. We can rewrite this as $n^2 - Cn - |C| > 0$. The roots of $n^2 - Cn - |C| = 0$ are $n = \frac{C \pm

\sqrt{C^2 + 4|C|}}{2}$. We want the positive root, so $n > \frac{C +

\sqrt{C^2 + 4|C|}}{2}$. Thus, we can choose $N = \frac{C +

\sqrt{C^2 + 4|C|}}{2}$. This ensures that for any n > N, $\frac{n^2}{n+sin n} > C$. This one uses the fact that sin(n) is bounded, which helps us make some useful approximations!

3.10. Proving the Limit of a Sum

Now, let's move on to proving a fundamental theorem about limits. We want to prove that if $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$, then $\lim_{n \to \infty} (a_n + b_n) = A + B$.

Proof:

Let $\epsilon > 0$ be given. Since $\lim_{n \to \infty} a_n = A$, there exists an integer $N_1$ such that for all n > N_1, $|a_n - A| < \frac{\epsilon}{2}$. Similarly, since $\lim_{n \to \infty} b_n = B$, there exists an integer $N_2$ such that for all n > N_2, $|b_n - B| < \frac{\epsilon}{2}$.

Now, let N = max(N_1, N_2). Then, for all n > N, we have both $|a_n - A| < \frac{\epsilon}{2}$ and $|b_n - B| < \frac{\epsilon}{2}$. We want to show that $|(a_n + b_n) - (A + B)| < \epsilon$ for all n > N.

We can write:

$|(a_n + b_n) - (A + B)| = |(a_n - A) + (b_n - B)| \leq |a_n - A| + |b_n - B|$ (by the triangle inequality)

Since n > N, we know that $|a_n - A| < \frac{\epsilon}{2}$ and $|b_n - B| < \frac{\epsilon}{2}$. Therefore:

$|(a_n + b_n) - (A + B)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Thus, for any $\epsilon > 0$, we have found an N such that for all n > N, $|(a_n + b_n) - (A + B)| < \epsilon$. This proves that $\lim_{n \to \infty} (a_n + b_n) = A + B|$.

In simpler terms, the limit of the sum is the sum of the limits! This is a super useful property when dealing with limits of sequences. The proof uses the epsilon-N definition of a limit and the triangle inequality, which are fundamental tools in real analysis. By carefully choosing N, we can ensure that the sum of the sequences gets arbitrarily close to the sum of their limits.

Conclusion

So there you have it! We successfully found values of N for different sequences to satisfy the condition $a_n > C$ and proved that the limit of the sum of two sequences is the sum of their limits. These are fundamental concepts in calculus and analysis, and understanding them is crucial for further studies in mathematics. Keep practicing, and you'll become a limit master in no time!