Divergence Or Convergence? Analyzing $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{(n+1)^3}$

Hey math enthusiasts! Let's dive into the fascinating world of series and tackle the question of whether the series $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{(n+1)^3}$ converges or diverges. We're going to use a handy tool called the nth term test (also known as the divergence test) to see what conclusions we can draw. This test is like a quick initial check to see if a series has a shot at converging. If the test fails, we can definitively say the series diverges. However, if the test passes, it doesn't necessarily mean the series converges; the test is inconclusive, and we'll need to use a different approach to figure it out. So, let's break down this series and see what the nth term test reveals! We will focus on understanding the behavior of the series by examining the terms as n approaches infinity. This is a fundamental concept in calculus, and understanding it is key to determining the convergence or divergence of a series. In this discussion, we will be using the nth term test to investigate the series $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{(n+1)^3}$. This will provide us with initial insight into the behavior of the series, and it's a critical first step in series analysis.

Understanding the Nth Term Test and Its Application

Alright, guys, let's get our heads around the nth term test. It's a straightforward test designed to help us determine if a series diverges. The core idea is this: if the terms of a series don't approach zero as n goes to infinity, then the series diverges. It's pretty intuitive, right? If the individual terms are not getting smaller and smaller, the sum can't possibly settle down to a finite value. Imagine adding a bunch of numbers together, and they're all consistently big; the total will just keep growing forever! Mathematically, the nth term test states: if $\lim_{n\to\infty} a_n \neq 0$, then the series $\sum_{n=1}^{\infty} a_n$ diverges. Here, $a_n$ represents the nth term of the series. The converse, however, isn't true. If $\lim_{n\to\infty} a_n = 0$, the test is inconclusive. This means the series might converge, but we'll need a more powerful test to confirm it. For this test to be useful, it must be well understood in the context of the problem, so let's get to know the series $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{(n+1)^3}$. Before proceeding with the nth term test, understanding the individual components of the series is crucial. We will investigate what happens to the nth term, $\frac{\sqrt{n}}{(n+1)^3}$, as n becomes extremely large. This understanding will allow us to apply the nth term test correctly and make an informed decision about the series' convergence or divergence. The limit will provide us with key insights into the series' long-term behavior. Understanding these behaviors is fundamental to series analysis.

Let's apply this test to our series. We have $a_n = \frac{\sqrt{n}}{(n+1)^3}$. To use the test, we need to find $\lim_{n\to\infty} a_n$.

Evaluating the Limit of the Series' Terms

Now, let's get down to the nitty-gritty and calculate the limit: $\lim_{n\to\infty} \frac{\sqrt{n}}{(n+1)^3}$. To evaluate this limit, we can use a little algebraic manipulation. The trick is often to divide both the numerator and denominator by the highest power of n present in the denominator. In our case, the highest power of n in the denominator (when expanded) is $n^3$. So, let's divide both the numerator and the denominator by $n^{3/2}$ (since $\sqrt{n} = n^{1/2}$ and we have $n^3$ in the denominator):

$\lim_{n\to\infty} \frac{\sqrt{n}}{(n+1)^3} = \lim_{n\to\infty} \frac{\frac{\sqrt{n}}{n^{3/2}}}{\frac{(n+1)^3}{n^{3/2}}} = \lim_{n\to\infty} \frac{\frac{1}{n^1}}{(1+\frac{1}{n})^3 n^{3/2-3}} = \lim_{n\to\infty} \frac{\frac{1}{n^1}}{(n+1)^3/n^{3}} = \lim_{n\to\infty} \frac{\frac{1}{n}}{(1+\frac{1}{n})^3 n^0}= \lim_{n\to\infty} \frac{\frac{1}{n}}{(1+\frac{1}{n})^3} = \lim_{n\to\infty} \frac{\frac{1}{n}}{1+3/n+3/n^2+1/n^3} = 0$

As n approaches infinity, $\frac{1}{n}$ approaches 0, and $(1 + \frac{1}{n})^3$ approaches 1. Therefore, the limit is 0. This manipulation simplifies the expression, making it easier to see how the limit behaves as n approaches infinity. It transforms the original expression into a form where we can directly apply limit properties and derive a conclusion.

Conclusion Based on the Nth Term Test

So, what does this limit calculation tell us, friends? Because $\lim_{n\to\infty} \frac{\sqrt{n}}{(n+1)^3} = 0$, the nth term test is inconclusive. It does not mean that the series converges or diverges. It just means that the nth term test doesn't provide enough information to determine the series' behavior. The nth term test is only useful for showing divergence, and the series could still either converge or diverge. We can't say for sure whether the series converges or diverges based solely on this test. We'll need to use a more sophisticated convergence test to determine whether the series converges or diverges. We have to employ a more advanced method to examine the convergence of the series, for example, the integral test or the comparison test. These tests will provide more insight into the series' behavior. It means that to fully understand the series, we need to step up our game and use more advanced techniques. Keep in mind that the nth term test is a good starting point, but it's not always the final answer!

To summarize:

• We used the nth term test.
• We found that the limit of the nth term of the series as n approaches infinity is 0.
• The nth term test is inconclusive.

So, the answer is: C. The test is inconclusive. We need to use a different test to determine whether the series converges or diverges. Let's keep exploring the fascinating world of series!