Series Convergence: Does $\frac{7}{\sqrt{x+1}\ln(\sqrt{x+1})}$ Converge?

Hey guys! Today, we're diving into a fascinating question about the convergence or divergence of a series. Specifically, we'll be looking at the series defined by $a = \frac{7}{\sqrt{x+1}\ln(\sqrt{x+1})}$. This is a classic problem that allows us to explore different techniques for determining the behavior of infinite series. So, buckle up and let's get started!

Understanding the Series and Initial Intuition

In this section, let's break down the series and try to get a feel for what might be happening. The series is given by $a = \frac{7}{\sqrt{x+1}\ln(\sqrt{x+1})}$. Before we jump into rigorous tests, let's take a moment to develop some intuition. We need to analyze the behavior of this expression as $x$ increases towards infinity. The numerator is a constant, 7, so that won't affect whether the series converges or diverges. Our focus should be on the denominator, which contains both a square root term, $\sqrt{x+1}$, and a logarithmic term, $\ln(\sqrt{x+1})$.

As $x$ grows larger, both $\sqrt{x+1}$ and $\ln(\sqrt{x+1})$ also increase, albeit at different rates. Square root functions grow slower than linear functions, but faster than logarithmic functions. Logarithmic functions, on the other hand, grow incredibly slowly. This interplay between the square root and the logarithm is crucial. Intuitively, since both terms are increasing, their product will also increase, causing the overall fraction to decrease. The crucial question is: does it decrease fast enough for the series to converge? Or does it decrease slowly enough that the series diverges? This is where more rigorous tests come into play.

Someone has already attempted to solve this using the integral test and arrived at the conclusion that the series diverges. This is a great starting point! The integral test is a powerful tool for determining convergence or divergence, especially when dealing with functions that are continuous, positive, and decreasing. The fact that the integral test suggests divergence gives us a strong hint. However, it's always good to have multiple lines of reasoning. Let's think about why divergence might be the case. If the function decreases very slowly, the area under the curve (which the integral test examines) might be infinite, indicating divergence. On the other hand, “looking at the series intuitively,” as the original question suggests, can sometimes be misleading. Our intuition needs to be carefully guided by mathematical principles. It’s possible that the initial intuition might not fully capture the subtle balance between the decreasing terms. This is why we need to delve deeper and potentially explore other convergence tests to confirm our findings. So, let's move on and rigorously explore the integral test and other methods to solidify our understanding.

Applying the Integral Test

Let's rigorously apply the integral test to the series. This is a crucial step in confirming the initial solution and understanding why the series diverges. The integral test states that if we have a continuous, positive, and decreasing function $f(x)$ on the interval $[1, \infty)$, then the infinite series $\sum_{x=1}^{\infty} f(x)$ converges if and only if the improper integral $\int_{1}^{\infty} f(x) dx$ converges. Conversely, if the integral diverges, the series also diverges. In our case, the function is $f(x) = \frac{7}{\sqrt{x+1}\ln(\sqrt{x+1})}$. We need to verify that this function is indeed continuous, positive, and decreasing for $x \geq 1$.

• Continuity: The function is continuous for $x > 0$ because both $\sqrt{x+1}$ and $\ln(\sqrt{x+1})$ are continuous in their respective domains, and their product is also continuous. The denominator is non-zero for $x > 0$, ensuring the function is continuous.

• Positivity: For $x \geq 1$, both $\sqrt{x+1}$ and $\ln(\sqrt{x+1})$ are positive, so their product is positive. The numerator is also positive, so the entire function is positive.

• Decreasing: To check if the function is decreasing, we can analyze its derivative. Let's find $f'(x)$. First, rewrite $f(x)$ as $f(x) = 7(\sqrt{x+1}\ln(\sqrt{x+1}))^{-1}$. Now, apply the chain rule and product rule:

  $f'(x) = -7(\sqrt{x+1}\ln(\sqrt{x+1}))^{-2} \cdot \frac{d}{dx} [\sqrt{x+1}\ln(\sqrt{x+1})]$

  We need to find the derivative of $\sqrt{x+1}\ln(\sqrt{x+1})$. Let $u = \sqrt{x+1}$ and $v = \ln(\sqrt{x+1}) = \frac{1}{2}\ln(x+1)$. Then, using the product rule:

  $\frac{d}{dx}(uv) = u'v + uv' = \frac{1}{2\sqrt{x+1}}\ln(\sqrt{x+1}) + \sqrt{x+1} \cdot \frac{1}{2(x+1)} = \frac{\ln(\sqrt{x+1})}{2\sqrt{x+1}} + \frac{1}{2\sqrt{x+1}}$

  So,

  $f'(x) = -7(\sqrt{x+1}\ln(\sqrt{x+1}))^{-2} \cdot \left( \frac{\ln(\sqrt{x+1})}{2\sqrt{x+1}} + \frac{1}{2\sqrt{x+1}} \right)$

  Since all terms within the parentheses are positive for $x \geq 1$, and we have a negative sign outside, $f'(x)$ is negative. This means the function is decreasing.

Now that we've verified the conditions for the integral test, let's evaluate the improper integral:

$\int_{1}^{\infty} \frac{7}{\sqrt{x+1}\ln(\sqrt{x+1})} dx$

To solve this, we can use a substitution. Let $u = \ln(\sqrt{x+1}) = \frac{1}{2}\ln(x+1)$. Then, $du = \frac{1}{2(x+1)}dx$, so $dx = 2(x+1)du$. Also, $\sqrt{x+1} = e^u$. When $x = 1$, $u = \ln(\sqrt{2})$, and as $x \to \infty$, $u \to \infty$. The integral becomes:

$\int_{\ln(\sqrt{2})}^{\infty} \frac{7}{e^u u} 2(x+1) du = 14 \int_{\ln(\sqrt{2})}^{\infty} \frac{e^{2u}}{e^u u} du = 14 \int_{\ln(\sqrt{2})}^{\infty} \frac{e^u}{u} du$

This integral is tricky to evaluate directly, but we can see that as $u$ goes to infinity, the integrand $\frac{e^u}{u}$ also goes to infinity. The integral diverges because the exponential function grows much faster than the linear function in the denominator. Therefore, since the integral diverges, by the integral test, the series $\sum_{x=1}^{\infty} \frac{7}{\sqrt{x+1}\ln(\sqrt{x+1})}$ also diverges. This confirms the initial solution obtained using the integral test.

Exploring Other Convergence Tests and Intuition Revisited

While the integral test clearly shows that the series diverges, let's take a moment to explore other convergence tests and revisit our intuition. This will give us a more comprehensive understanding of why this series behaves the way it does. One test we could consider is the comparison test. The comparison test is useful when we can compare our given series to another series whose convergence or divergence is already known. However, finding a suitable series to compare with can sometimes be challenging. In our case, we might try comparing with a $p$-series or a simpler logarithmic series, but it's not immediately obvious which comparison would be most effective.

Another test is the limit comparison test, which is often more flexible than the direct comparison test. The limit comparison test states that if we have two series $\sum a_n$ and $\sum b_n$ with positive terms, and if $\lim_{n \to \infty} \frac{a_n}{b_n} = c$, where $c$ is a finite number greater than zero, then either both series converge or both diverge. To apply this test, we need to choose a series $b_n$ to compare with. Let's consider comparing our series with $b_n = \frac{1}{n}$, which we know diverges (it's the harmonic series). If we let $a_x = \frac{7}{\sqrt{x+1}\ln(\sqrt{x+1})}$, we would need to evaluate the limit:

$\lim_{x \to \infty} \frac{a_x}{b_x} = \lim_{x \to \infty} \frac{\frac{7}{\sqrt{x+1}\ln(\sqrt{x+1})}}{\frac{1}{x}} = \lim_{x \to \infty} \frac{7x}{\sqrt{x+1}\ln(\sqrt{x+1})}$

This limit is not immediately obvious, and we might need to use L'Hôpital's rule to evaluate it. However, the derivatives involved can get quite messy. So, while the limit comparison test is a viable option, it might not be the most straightforward in this case.

Let's revisit our intuition. We noted earlier that the denominator $\sqrt{x+1}\ln(\sqrt{x+1})$ increases as $x$ increases, but it does so relatively slowly. The square root function grows more slowly than a linear function, and the logarithm grows even more slowly. This slow growth in the denominator means that the terms of the series don't decrease rapidly enough for the series to converge. The integral test captured this slow decay by showing that the area under the curve is infinite. In essence, the terms of the series, while getting smaller, don't get small quickly enough to sum to a finite value. This is a key concept in understanding why certain series diverge. Think about it like this: even if you're adding smaller and smaller pieces, if you add infinitely many of them and they don't shrink fast enough, you'll end up with an infinite sum.

Conclusion: The Series Diverges

In conclusion, through the rigorous application of the integral test, we've definitively shown that the series given by $a = \frac{7}{\sqrt{x+1}\ln(\sqrt{x+1})}$ diverges. We also explored other potential convergence tests, such as the comparison test and the limit comparison test, to broaden our understanding. Revisiting our initial intuition, we recognize that the slow growth of the denominator plays a crucial role in the divergence. The terms of the series decrease, but not rapidly enough to yield a finite sum. This exercise highlights the importance of both rigorous mathematical tools and intuitive reasoning in analyzing the behavior of infinite series. It's crucial to not only apply the tests correctly but also to understand why a series converges or diverges. I hope this detailed analysis has helped you guys grasp the concepts involved and appreciate the fascinating world of infinite series! Thanks for joining me on this exploration!