Evaluate The Following Infinite Series:$\[ \sum_{n=1}^{\infty} N^2\left(\frac{5}{6}\right)^n \\]

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Introduction

Infinite series are a fundamental concept in mathematics, and they have numerous applications in various fields, including physics, engineering, and economics. The given infinite series is a power series, which is a series of the form ∑n=0∞anxn\sum_{n=0}^{\infty} a_n x^n. In this case, the series is ∑n=1∞n2(56)n\sum_{n=1}^{\infty} n^2\left(\frac{5}{6}\right)^n. The objective of this article is to evaluate this infinite series and provide a detailed analysis of its convergence.

Background

Before we dive into the evaluation of the series, let's briefly discuss the concept of infinite series and their convergence. An infinite series is a sum of an infinite number of terms, and it can be written in the form ∑n=1∞an\sum_{n=1}^{\infty} a_n. The series is said to converge if the sum of the terms approaches a finite limit as the number of terms increases without bound. On the other hand, if the sum of the terms does not approach a finite limit, the series is said to diverge.

The Series

The given series is ∑n=1∞n2(56)n\sum_{n=1}^{\infty} n^2\left(\frac{5}{6}\right)^n. This is a power series with the general term an=n2(56)na_n = n^2\left(\frac{5}{6}\right)^n. To evaluate this series, we can use the formula for the sum of a geometric series, which is given by ∑n=0∞arn=a1−r\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}, where aa is the first term and rr is the common ratio.

Derivation

To derive the formula for the sum of the given series, we can start by writing the series as:

∑n=1∞n2(56)n=∑n=1∞n2(56)n⋅65⋅65\sum_{n=1}^{\infty} n^2\left(\frac{5}{6}\right)^n = \sum_{n=1}^{\infty} n^2\left(\frac{5}{6}\right)^n \cdot \frac{6}{5} \cdot \frac{6}{5}

Now, we can use the formula for the sum of a geometric series to rewrite the series as:

∑n=1∞n2(56)n=65⋅65∑n=1∞n2(56)n\sum_{n=1}^{\infty} n^2\left(\frac{5}{6}\right)^n = \frac{6}{5} \cdot \frac{6}{5} \sum_{n=1}^{\infty} n^2\left(\frac{5}{6}\right)^n

Simplification

We can simplify the expression by canceling out the common factor of 65\frac{6}{5}:

∑n=1∞n2(56)n=3625∑n=1∞n2(56)n\sum_{n=1}^{\infty} n^2\left(\frac{5}{6}\right)^n = \frac{36}{25} \sum_{n=1}^{\infty} n^2\left(\frac{5}{6}\right)^n

Convergence

To determine whether the series converges, we can use the ratio test. The ratio test states that if the limit of the ratio of consecutive terms is less than 1, the series converges. In this case, the ratio of consecutive terms is:

an+1an=(n+1)2(56)n+1n2(56)n\frac{a_{n+1}}{a_n} = \frac{(n+1)^2\left(\frac{5}{6}\right)^{n+1}}{n^2\left(\frac{5}{6}\right)^n}

Simplification

We can simplify the expression by canceling out the common factor of (56)n\left(\frac{5}{6}\right)^n:

an+1an=(n+1)2n2⋅56\frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{n^2} \cdot \frac{5}{6}

Limit

To determine whether the series converges, we can take the limit of the ratio of consecutive terms as nn approaches infinity:

lim⁡n→∞an+1an=lim⁡n→∞(n+1)2n2⋅56\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(n+1)^2}{n^2} \cdot \frac{5}{6}

Simplification

We can simplify the expression by canceling out the common factor of n2n^2:

lim⁡n→∞an+1an=lim⁡n→∞(n+1n)2⋅56\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^2 \cdot \frac{5}{6}

Limit

To determine whether the series converges, we can take the limit of the expression as nn approaches infinity:

lim⁡n→∞(n+1n)2⋅56=(lim⁡n→∞n+1n)2⋅56\lim_{n \to \infty} \left(\frac{n+1}{n}\right)^2 \cdot \frac{5}{6} = \left(\lim_{n \to \infty} \frac{n+1}{n}\right)^2 \cdot \frac{5}{6}

Simplification

We can simplify the expression by canceling out the common factor of nn:

(lim⁡n→∞n+1n)2⋅56=(lim⁡n→∞n+1n)2⋅56\left(\lim_{n \to \infty} \frac{n+1}{n}\right)^2 \cdot \frac{5}{6} = \left(\lim_{n \to \infty} \frac{n+1}{n}\right)^2 \cdot \frac{5}{6}

Limit

To determine whether the series converges, we can take the limit of the expression as nn approaches infinity:

(lim⁡n→∞n+1n)2⋅56=12⋅56\left(\lim_{n \to \infty} \frac{n+1}{n}\right)^2 \cdot \frac{5}{6} = 1^2 \cdot \frac{5}{6}

Simplification

We can simplify the expression by canceling out the common factor of 121^2:

12⋅56=561^2 \cdot \frac{5}{6} = \frac{5}{6}

Conclusion

Since the limit of the ratio of consecutive terms is less than 1, the series converges.

Final Answer

The final answer is 30(1−56)3\boxed{\frac{30}{(1-\frac{5}{6})^3}}.

Derivation

To derive the final answer, we can use the formula for the sum of a geometric series, which is given by ∑n=0∞arn=a1−r\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}, where aa is the first term and rr is the common ratio.

Simplification

We can simplify the expression by canceling out the common factor of 56\frac{5}{6}:

30(1−56)3=30(16)3\frac{30}{(1-\frac{5}{6})^3} = \frac{30}{(\frac{1}{6})^3}

Simplification

We can simplify the expression by canceling out the common factor of 16\frac{1}{6}:

30(16)3=30⋅63\frac{30}{(\frac{1}{6})^3} = 30 \cdot 6^3

Simplification

We can simplify the expression by canceling out the common factor of 636^3:

30⋅63=30⋅21630 \cdot 6^3 = 30 \cdot 216

Simplification

We can simplify the expression by canceling out the common factor of 3030:

30⋅216=648030 \cdot 216 = 6480

Conclusion

The final answer is 6480\boxed{6480}.

Final Answer

The final answer is 6480\boxed{6480}.

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Introduction

In the previous article, we evaluated the infinite series ∑n=1∞n2(56)n\sum_{n=1}^{\infty} n^2\left(\frac{5}{6}\right)^n. In this article, we will answer some frequently asked questions related to the evaluation of infinite series.

Q: What is an infinite series?

A: An infinite series is a sum of an infinite number of terms. It can be written in the form ∑n=1∞an\sum_{n=1}^{\infty} a_n, where ana_n is the general term of the series.

Q: What is the difference between a convergent and a divergent series?

A: A convergent series is a series that approaches a finite limit as the number of terms increases without bound. A divergent series, on the other hand, is a series that does not approach a finite limit as the number of terms increases without bound.

Q: How do I determine whether a series converges or diverges?

A: To determine whether a series converges or diverges, you can use various tests, such as the ratio test, the root test, or the integral test.

Q: What is the ratio test?

A: The ratio test is a test used to determine whether a series converges or diverges. It states that if the limit of the ratio of consecutive terms is less than 1, the series converges.

Q: What is the root test?

A: The root test is a test used to determine whether a series converges or diverges. It states that if the limit of the nth root of the general term is less than 1, the series converges.

Q: What is the integral test?

A: The integral test is a test used to determine whether a series converges or diverges. It states that if the integral of the general term is finite, the series converges.

Q: How do I evaluate a series using the formula for the sum of a geometric series?

A: To evaluate a series using the formula for the sum of a geometric series, you need to identify the first term and the common ratio of the series. Then, you can use the formula ∑n=0∞arn=a1−r\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} to evaluate the series.

Q: What is the formula for the sum of a geometric series?

A: The formula for the sum of a geometric series is ∑n=0∞arn=a1−r\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}, where aa is the first term and rr is the common ratio.

Q: How do I determine the first term and the common ratio of a series?

A: To determine the first term and the common ratio of a series, you need to examine the general term of the series. The first term is the coefficient of the first power of the variable, and the common ratio is the ratio of consecutive terms.

Q: What are some common types of series?

A: Some common types of series include geometric series, arithmetic series, and power series.

Q: What is a geometric series?

A: A geometric series is a series of the form ∑n=0∞arn\sum_{n=0}^{\infty} ar^n, where aa is the first term and rr is the common ratio.

Q: What is an arithmetic series?

A: An arithmetic series is a series of the form ∑n=0∞(a+nd)\sum_{n=0}^{\infty} (a+nd), where aa is the first term and dd is the common difference.

Q: What is a power series?

A: A power series is a series of the form ∑n=0∞anxn\sum_{n=0}^{\infty} a_n x^n, where ana_n is the general term and xx is the variable.

Q: How do I use a power series to approximate a function?

A: To use a power series to approximate a function, you need to find the power series representation of the function. Then, you can use the power series to approximate the function at a given point.

Q: What are some common applications of infinite series?

A: Some common applications of infinite series include physics, engineering, and economics.

Q: How do I use infinite series in physics?

A: Infinite series are used in physics to model various phenomena, such as the motion of objects, the behavior of electrical circuits, and the properties of materials.

Q: How do I use infinite series in engineering?

A: Infinite series are used in engineering to design and analyze various systems, such as electrical circuits, mechanical systems, and control systems.

Q: How do I use infinite series in economics?

A: Infinite series are used in economics to model various economic phenomena, such as the behavior of markets, the growth of economies, and the distribution of wealth.

Conclusion

In this article, we have answered some frequently asked questions related to the evaluation of infinite series. We have discussed the concept of infinite series, the difference between convergent and divergent series, and various tests used to determine whether a series converges or diverges. We have also discussed the formula for the sum of a geometric series and how to use it to evaluate a series. Finally, we have discussed some common applications of infinite series in physics, engineering, and economics.