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Convergence and Divergence of a Given Series: A Mathematical Analysis

In mathematics, the convergence and divergence of a series are crucial concepts that help us understand the behavior of infinite sequences. A series is said to converge if its sum approaches a finite value as the number of terms increases, while it diverges if the sum grows indefinitely. In this article, we will analyze the convergence and divergence of a given series, which is represented as:

$$\sum_{n=2}^{\infty} \frac{x^n}{n^2} + \frac{x^{3n}}{n^2}$$

where $x$ is a real number. We will prove that this series converges when $x < 1$ and diverges when $x \geq 1$.

The given series is a combination of two separate series, each with its own convergence properties. The first series is $\sum_{n=2}^{\infty} \frac{x^n}{n^2}$, and the second series is $\sum_{n=2}^{\infty} \frac{x^{3n}}{n^2}$. We will analyze each series separately and then combine the results to determine the convergence or divergence of the given series.

The first series is $\sum_{n=2}^{\infty} \frac{x^n}{n^2}$. To determine its convergence or divergence, we can use the ratio test. The ratio test states that a series $\sum_{n=1}^{\infty} a_n$ converges if the limit of $\left| \frac{a_{n+1}}{a_n} \right|$ as $n$ approaches infinity is less than 1.

Let's apply the ratio test to the first series:

$$\lim_{n \to \infty} \left| \frac{\frac{x^{n+1}}{(n+1)^2}}{\frac{x^n}{n^2}} \right| = \lim_{n \to \infty} \left| \frac{x^{n+1} n^2}{x^n (n+1)^2} \right| = \lim_{n \to \infty} \left| \frac{x n^2}{(n+1)^2} \right| = \left| x \right|$$

Since the limit is $\left| x \right|$, the series converges if $\left| x \right| < 1$ and diverges if $\left| x \right| \geq 1$. Therefore, the first series converges when $x < 1$ and diverges when $x \geq 1$.

The second series is $\sum_{n=2}^{\infty} \frac{x^{3n}}{n^2}$. We can also use the ratio test to determine its convergence or divergence:

$$\lim_{n \to \infty} \left| \frac{\frac{x^{3(n+1)}}{(n+1)^2}}{\frac{x^{3n}}{n^2}} \right| = \lim_{n \to \infty} \left| \frac{x^{3(n+1)} n^2}{x^{3n} (n+1)^2} \right| = \lim_{n \to \infty} \left| \frac{x^{3n} x^3 n^2}{x^{3n} (n+1)^2} \right| = \lim_{n \to \infty} \left| \frac{x^3 n^2}{(n+1)^2} \right| = \left| x^3 \right|$$

Since the limit is $\left| x^3 \right|$, the series converges if $\left| x^3 \right| < 1$ and diverges if $\left| x^3 \right| \geq 1$. Therefore, the second series converges when $x^3 < 1$ and diverges when $x^3 \geq 1$.

Now that we have analyzed each series separately, we can combine the results to determine the convergence or divergence of the given series. The given series is a combination of the two series, so we need to consider the convergence or divergence of both series simultaneously.

From the analysis of the first series, we know that it converges when $x < 1$ and diverges when $x \geq 1$. From the analysis of the second series, we know that it converges when $x^3 < 1$ and diverges when $x^3 \geq 1$.

Since $x^3 < 1$ when $x < 1$, the second series converges when $x < 1$. Therefore, the given series converges when $x < 1$.

On the other hand, when $x \geq 1$, the first series diverges. Therefore, the given series diverges when $x \geq 1$.

In conclusion, we have analyzed the convergence and divergence of a given series, which is represented as:

$$\sum_{n=2}^{\infty} \frac{x^n}{n^2} + \frac{x^{3n}}{n^2}$$

where $x$ is a real number. We have proved that this series converges when $x < 1$ and diverges when $x \geq 1$. The analysis of the series was done by considering the convergence or divergence of two separate series, each with its own convergence properties. The results were then combined to determine the convergence or divergence of the given series.

• [1] Rudin, W. (1976). Principles of Mathematical Analysis. McGraw-Hill.
• [2] Apostol, T. M. (1974). Mathematical Analysis. Addison-Wesley.
• [3] Spivak, M. (1965). Calculus. W.A. Benjamin.

In the future, we can extend this analysis to more complex series and explore the convergence or divergence of other types of series. We can also investigate the properties of the given series, such as its rate of convergence or divergence, and its behavior under different conditions.

The following code can be used to verify the convergence or divergence of the given series:

import numpy as np
def series(x, n):
return np.sum([xi / i2 + x**(3*i) / i**2 for i in range(2, n)])
x = 0.5
n = 1000
result = series(x, n)
print(result)

This code defines a function series that calculates the sum of the given series up to a certain number of terms n. The function takes two arguments: x, the value of the variable, and n, the number of terms. The code then calls the function with x = 0.5 and n = 1000 and prints the result.

Note that this code is for illustrative purposes only and is not intended to be used for actual calculations. The convergence or divergence of the series depends on the value of x, and the code should be modified accordingly.
Q&A: Convergence and Divergence of a Given Series

In our previous article, we analyzed the convergence and divergence of a given series, which is represented as:

$$\sum_{n=2}^{\infty} \frac{x^n}{n^2} + \frac{x^{3n}}{n^2}$$

where $x$ is a real number. We proved that this series converges when $x < 1$ and diverges when $x \geq 1$. In this article, we will answer some frequently asked questions about the convergence and divergence of this series.

Q: What is the significance of the value $x=1$ in the convergence and divergence of the series?

A: The value $x=1$ is a critical point in the convergence and divergence of the series. When $x < 1$, the series converges, and when $x \geq 1$, the series diverges. This is because the terms of the series behave differently when $x$ is less than or greater than 1.

Q: Can you explain why the series converges when $x < 1$?

A: Yes, the series converges when $x < 1$ because the terms of the series decrease rapidly as $n$ increases. This is due to the fact that the exponent $x^n$ decreases as $n$ increases when $x < 1$. As a result, the sum of the terms of the series approaches a finite value as $n$ approaches infinity.

Q: Can you explain why the series diverges when $x \geq 1$?

A: Yes, the series diverges when $x \geq 1$ because the terms of the series increase rapidly as $n$ increases. This is due to the fact that the exponent $x^n$ increases as $n$ increases when $x \geq 1$. As a result, the sum of the terms of the series grows indefinitely as $n$ approaches infinity.

Q: How can we determine whether a series converges or diverges?

A: There are several methods to determine whether a series converges or diverges, including the ratio test, the root test, and the integral test. The ratio test is a common method used to determine the convergence or divergence of a series.

Q: What is the ratio test?

A: The ratio test is a method used to determine the convergence or divergence of a series. It involves calculating the limit of the ratio of consecutive terms of the series as $n$ approaches infinity. If the limit is less than 1, the series converges. If the limit is greater than 1, the series diverges.

Q: Can you provide an example of how to use the ratio test to determine the convergence or divergence of a series?

A: Yes, here is an example of how to use the ratio test to determine the convergence or divergence of the series:

$$\sum_{n=1}^{\infty} \frac{1}{n^2}$$

To use the ratio test, we need to calculate the limit of the ratio of consecutive terms of the series as $n$ approaches infinity:

$$\lim_{n \to \infty} \left| \frac{\frac{1}{(n+1)^2}}{\frac{1}{n^2}} \right| = \lim_{n \to \infty} \left| \frac{n^2}{(n+1)^2} \right| = 1$$

Since the limit is 1, the series diverges.

Q: What is the root test?

A: The root test is a method used to determine the convergence or divergence of a series. It involves calculating the limit of the $n$th root of the $n$th term of the series as $n$ approaches infinity. If the limit is less than 1, the series converges. If the limit is greater than 1, the series diverges.

Q: Can you provide an example of how to use the root test to determine the convergence or divergence of a series?

A: Yes, here is an example of how to use the root test to determine the convergence or divergence of the series:

$$\sum_{n=1}^{\infty} \frac{1}{n}$$

To use the root test, we need to calculate the limit of the $n$th root of the $n$th term of the series as $n$ approaches infinity:

$$\lim_{n \to \infty} \sqrt[n]{\frac{1}{n}} = \lim_{n \to \infty} \frac{1}{n^{\frac{1}{n}}} = 1$$

Since the limit is 1, the series diverges.

In this article, we have answered some frequently asked questions about the convergence and divergence of a given series. We have explained the significance of the value $x=1$ in the convergence and divergence of the series, and we have provided examples of how to use the ratio test and the root test to determine the convergence or divergence of a series. We hope that this article has been helpful in understanding the convergence and divergence of series.