Problem 4 - The Calculus with Analytic Geometry

Question

Find the interval of convergence of the given power series.\(\sum_{n=1}^{+\infty}(-1)^{n} \frac{x^{2 n}}{(2 n) !}\)

Step-by-Step Solution

Answer

The interval of convergence is \((-\infty, \infty)\).

Step 1 - Identify the general term of the series

The given power series is \(\sum_{n=1}^{+\infty}(-1)^{n} \frac{x^{2n}}{(2n)!} \). The general term of this series is \(-1)^{n} \frac{x^{2n}}{(2n)!} \).

Step 2 - Apply the Ratio Test

To find the interval of convergence, apply the Ratio Test. The Ratio Test uses the limit: \(L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \), where \(a_n = (-1)^n \frac{x^{2n}}{(2n)!} \).\Calculate the ratio: \(\frac{a_{n+1}}{a_n} = \frac{(-1)^{n+1} \frac{x^{2(n+1)}}{(2(n+1))!}}{(-1)^{n} \frac{x^{2n}}{(2n)!}} \= \frac{(-1)^{n+1} x^{2(n+1)}}{(2(n+1))!} \cdot \frac{(2n)!}{(-1)^n x^{2n}}\ = \left| x^2 \cdot \frac{1}{(2n+1)(2n+2)} \right| \= \left| x^2 \right| \cdot \frac{1}{(2n+1)(2n+2)} \).

Step 3 - Evaluate the limit

Taking the limit as \(n \to \infty\), we get: \(\lim_{{n \to \infty}} \left| x^2 \cdot \frac{1}{(2n+1)(2n+2)} \right| = x^2 \cdot \lim_{{n \to \infty}} \frac{1}{(2n+1)(2n+2)} = 0 \).

Step 4 - Determine the interval of convergence

Since the limit found in Step 3 is always 0 for any \(x\), it is less than 1 and the Ratio Test indicates that the series converges for all \(x\). Hence the interval of convergence is \((-\infty, \infty)\).

Key Concepts

Power Series

A power series is a sum of terms of the form \(a_n x^n \), where \(a_n\) are coefficients and \(x\) is a variable.
Power series are used to represent functions as infinite polynomials.
They are very useful because they can approximate functions to a high degree of accuracy.
The general form of a power series is: \[ \sum_{n=0}^{\infty} a_n x^n \].
The series in our exercise can be written as: \[ \sum_{n=1}^{\infty}(-1)^n \frac{x^{2n}}{(2n)!} \].
Each term in the series \((-1)^n \frac{x^{2n}}{(2n)!}\) includes a coefficient \((-1)^n \), a power of \(x\), and a factorial in the denominator.

Ratio Test

The Ratio Test helps determine the convergence of a series.
It involves calculating the limit of the ratio of consecutive terms.
The Ratio Test formula is: \ L = \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| \.
Based on the value of \(L\):

• If \(L < 1\), the series converges.
• If \(L > 1\), the series diverges.
• If \(L = 1\), the test is inconclusive.

In the given exercise, we used the Ratio Test on the series \(\sum_{n=1}^{ewline+ewlineinfty}(-1)^{n} \frac{x^{2n}}{(2n)!}\).
We evaluated \ \frac{a_{n+1}}{a_n} \ and found the limit to determine the convergence.

Series Convergence

Series convergence means the terms of the series approach a final value as \(n\) goes to infinity.
For a series \[ \sum_{n=0}^{\infty} a_n x^n \], if the sequence of partial sums converges to a limit, the series converges.
Convergence depends on the values of \(x\) in the series.
The interval of convergence is the set of all \(x\) values for which the series converges.

In our exercise, after performing the Ratio Test, we found: \ \lim_{n \rightarrow \infty} \left| x^2 \cdot \frac{1}{(2n+1)(2n+2)\right| = 0.
Since this limit is always less than 1 for all \(x\), the series converges for any real number.
Therefore, the interval of convergence is \((-\infty, \infty)\).

Limit Evaluation

Limit evaluation is finding the value that a function or sequence approaches as the input approaches some value.
It's often used in tests for series convergence, like the Ratio Test.
To evaluate limits, we may use algebraic manipulation, L'Hôpital's rule, or other methods.

In the provided solution, we evaluated the limit:
\[ \lim_{n \rightarrow \infty} \left| x^2 \cdot \frac{1}{(2n+1)(2n+2)\right| \].

When \(n\) becomes very large, \( \frac{1}{(2n+1)(2n+2)} \) approaches 0, making the entire limit 0.
This result helped us to conclude that the series converges for all \(x\).