Series Convergence: Math Problems Solved

Hey Plastik Magazine readers! Let's dive into some serious math today. We're going to break down the convergence or divergence of three series problems. Don't worry, it's not as scary as it sounds. We'll go step-by-step, making sure everything is super clear and easy to follow. Get ready to flex those math muscles!

n) Examining the Series: $\sum_{n=1}^{\infty} \frac{6 n-7}{\sqrt{4 n^2+5 n+6}}$

Alright, guys, let's tackle this first problem: $\sum_{n=1}^{\infty} \frac{6 n-7}{\sqrt{4 n^2+5 n+6}}$. The key here is to figure out whether this series converges (meaning it approaches a finite value) or diverges (meaning it goes off to infinity or doesn't settle down). One of the best initial tests to use is the Divergence Test, also known as the nth-Term Test. This test is super simple: if the limit of the general term of the series as n approaches infinity is not equal to zero, then the series diverges. Plain and simple. Let's see how that works for our series.

First, we need to find the limit of the general term, which is $\frac{6n - 7}{\sqrt{4n^2 + 5n + 6}}$. As n gets incredibly large, we can ignore the smaller terms (-7, 5n, and 6) because they become insignificant compared to the dominant terms (6n and 4n^2). So, we can approximate our general term as $\frac{6n}{\sqrt{4n^2}}$.

Now, simplifying $\sqrt{4n^2}$ gives us 2n. Therefore, our approximated term becomes $\frac{6n}{2n}$, which simplifies to 3. So, the limit of the general term as n approaches infinity is 3. Since this limit is not equal to 0, the Divergence Test tells us that our series, $\sum_{n=1}^{\infty} \frac{6 n-7}{\sqrt{4 n^2+5 n+6}}$, diverges. That's it! We've successfully determined the behavior of this series. The Divergence Test is a powerful tool to quickly identify whether a series is even capable of converging. If the terms aren't approaching zero, it's a guaranteed divergence. Easy peasy, right?

Let's go into more detail. The Divergence Test is based on a fundamental concept: for a series to converge, the individual terms must get infinitesimally small. If they don't, the sum can't settle down to a finite value. In our series, the terms don't shrink toward zero. Instead, they approach a constant value, which means the sum will keep growing, never converging.

To make this even clearer, we can consider the behavior of the numerator and denominator separately. The numerator, 6n - 7, grows linearly with n. The denominator, $\sqrt{4n^2 + 5n + 6}$, also grows, but it grows at a rate proportional to n (because of the square root). As n gets huge, the numerator and denominator behave similarly, so the fraction approaches a constant value (in this case, 3). Because the terms don't shrink towards zero, the series diverges.

We could also use the Limit Comparison Test. To do this, we compare our series to a known divergent series. We can compare to $\frac{6n}{2n}$, which we already know simplifies to 3. Therefore, because the limit is finite and non-zero, this would further confirm the divergence.

Therefore, we have demonstrated through the Divergence Test and reasoning about the behavior of the terms that this series diverges. Awesome job, everyone!

o) Analyzing the Series: $\sum_{n=1}^{\infty} \frac{2-3 e^{4 n}}{9 n+8}$

Alright, let's move on to the next problem: $\sum_{n=1}^{\infty} \frac{2-3 e^{4 n}}{9 n+8}$. We'll use the same basic approach as before: use the Divergence Test! It's generally a great starting point for checking series convergence or divergence. If the limit of the general term as n approaches infinity is not zero, we know the series diverges. Let's get to it!

Our general term is $\frac{2-3e^{4n}}{9n+8}$. As n gets super large, the exponential term, $e^{4n}$, will dominate the other terms. The constant '2' and the linear term '9n' in the denominator become insignificant. So, we can approximate our general term by looking at the limit of $\frac{-3e^{4n}}{9n}$ as n approaches infinity.

Here's where things get interesting! Exponential functions grow much faster than linear functions. As n increases, $e^{4n}$ grows incredibly quickly, while 9n grows linearly. Because of this, the numerator, $-3e^{4n}$, will grow to become infinitely large (or, in this case, infinitely negative). The denominator will also grow, but the numerator will always grow much faster. So the fraction tends towards infinity. More specifically, we would rewrite as:

$\lim_{n\to\infty} \frac{2-3e^{4n}}{9n+8} = \lim_{n\to\infty} \frac{-3e^{4n}}{9n}$.

We know that the exponential grows much faster than linear, which leads us to conclude that:

$\lim_{n\to\infty} \frac{-3e^{4n}}{9n} = -\infty$

As the limit of the general term is not zero (it's actually negative infinity), the Divergence Test tells us that our series, $\sum_{n=1}^{\infty} \frac{2-3 e^{4 n}}{9 n+8}$, diverges. This means the series does not approach a finite value, but instead tends toward negative infinity.

To really drive this home, let's consider another way to think about it. As n increases, the term $e^{4n}$ becomes huge very quickly. Since it's multiplied by -3 in the numerator, the whole numerator becomes a very large negative number. The denominator is a linear function of n, but it will never 'catch up' to the exponential in the numerator. The overall fraction continues to decrease without bound, confirming our divergence conclusion.

We could use other tests to confirm, but the Divergence Test is usually the easiest way to solve the problem and get to the answer quickly. Also, notice that any test that involves comparing would lead to the same conclusion, as the exponential function will always dominate any polynomial function in the denominator.

Great work, everyone! You’re getting the hang of this.

p) Investigating the Series: $\sum_{n=1}^{\infty} \frac{2 n^5}{\ln \left(7 n^8+1\right)}$

Last but not least, let's crack this series: $\sum_{n=1}^{\infty} \frac{2 n^5}{\ln \left(7 n^8+1\right)}$. Let's see if we can tame this beast. Once again, we'll start with the Divergence Test. It's our trusty friend in these scenarios. Let's calculate the limit of the general term as n approaches infinity.

Our general term is $\frac{2n^5}{\ln(7n^8 + 1)}$. As n gets really, really large, we can think about the behavior of the numerator and denominator separately. The numerator, $2n^5$, is a polynomial function that grows at a rate proportional to $n^5$. The denominator is a logarithmic function, $\ln(7n^8 + 1)$. We know that logarithmic functions grow, but they grow very slowly compared to polynomial functions. The $n^8$ inside the logarithm will dominate the '+1', so we can approximate the denominator as $\ln(7n^8)$.

Now, using the properties of logarithms, we can simplify $\ln(7n^8)$ as $\ln(7) + 8\ln(n)$. So, our general term can be approximated as $\frac{2n^5}{\ln(7) + 8\ln(n)}$. As n goes to infinity, the numerator ($2n^5$) grows much faster than the denominator ($ln(7) + 8\ln(n)$). This is because polynomial functions always outpace logarithmic functions as n becomes very large.

Therefore, the limit of $\frac{2n^5}{\ln(7) + 8\ln(n)}$ as n approaches infinity is infinity. Since the limit is not zero, our series diverges. We can officially say that the series $\sum_{n=1}^{\infty} \frac{2 n^5}{\ln \left(7 n^8+1\right)}$ diverges. Another win for our team!

To put a finer point on this, think about the race between the numerator and the denominator. The numerator, $2n^5$, increases rapidly. The denominator, because it has a logarithm, grows slowly. Because the numerator wins the race by increasing faster, the whole fraction goes to infinity. We know this behavior because we understand the growth rates of polynomials and logarithms. Polynomials always outpace logarithms.

We could also use the Limit Comparison Test here if we wanted. By comparing with $n^2$, we could show that the limit of the comparison is infinite, thereby confirming divergence. However, the Divergence Test is perfectly sufficient for this problem.

In summary, because the general term does not approach zero as n goes to infinity, the series must diverge. Congratulations!

And that wraps it up, folks! We've successfully examined three series for convergence or divergence. Keep practicing, and you'll become a series master in no time! Until next time, keep those math muscles flexed!