Chapter 8: Problem 51
Find the sum of the given series in closed form. State the radius of convergence \(R\). \(\sum_{n=1}^{\infty} n x^{2 n+1}\)
Short Answer
Expert verified
The closed form of the series is \( \frac{x^5}{(1-x^2)^2} \) and the radius of convergence is 1.
Step by step solution
01
Identify the Type of Series
The given series is \( \sum_{n=1}^{\infty} n x^{2n+1} \). This series is a power series, which can be represented as \( a_n x^n \) where \( a_n = n \). The series consists of terms in odd powers of \( x \).
02
Simplify the Series
Observe that the series can be rewritten as \( \sum_{n=1}^{\infty} n \cdot x^{2n+1} = x^3 \sum_{n=1}^{\infty} n (x^2)^n \). The series inside the summation, \( \sum_{n=1}^{\infty} n (x^2)^n \), is a geometric series with the first term \( a = x^2 \) and common ratio \( r = x^2 \).
03
Apply Formula for Sum of Geometric Series
For a geometric series of the form \( \sum_{n=0}^{\infty} a r^n = \frac{a}{1-r} \), we first note the series here actually starts at \( n=1 \), so we must adjust by subtracting the term for \( n=0 \), which gives \( \sum_{n=1}^{\infty} n (x^2)^n = \frac{x^2}{(1 - x^2)^2} \).
04
Calculate Closed Form of the Original Series
Utilizing the result from Step 3, the original series \( x^3 \sum_{n=1}^{\infty} n (x^2)^n \) becomes \( x^3 \cdot \frac{x^2}{(1-x^2)^2} = \frac{x^5}{(1-x^2)^2} \). This is the closed form of the series.
05
Determine Radius of Convergence
The power series \( \sum_{n=1}^{\infty} n (x^2)^n \) converges when \( |x^2| < 1 \), implying \( |x| < 1 \). Thus, the radius of convergence \( R \) is given by \( R = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Series
A geometric series is a series of terms with a constant ratio between successive terms. Imagine you are stacking blocks, and you add the same amount of blocks each time you stack. That's like a geometric series where you multiply by a fixed number, known as the common ratio, with each step. In the given series, when we rewrite it as \( \sum_{n=1}^{\infty} n (x^2)^n \), we recognize this as a geometric series.
This is because each term can be generated by multiplying a fixed first term by a continually increasing power of \( x^2 \).
This is because each term can be generated by multiplying a fixed first term by a continually increasing power of \( x^2 \).
- The first term, \( a \), is \( x^2 \).
- The common ratio, \( r \), is also \( x^2 \).
Radius of Convergence
The radius of convergence of a power series tells us the value of \( x \) for which the series converges. Picture the radius of a circle; it tells how far you can go out from the center before you can't anymore. For a power series like \( \sum_{n=1}^{\infty} n (x^2)^n \), convergence occurs when the absolute value of \( x^2 \) is less than 1.
This condition, \(|x^2| < 1\), simplifies to \(|x| < 1\). It's like saying as long as \( x \) stays within this bound, the series behaves well and finds a sum.
This condition, \(|x^2| < 1\), simplifies to \(|x| < 1\). It's like saying as long as \( x \) stays within this bound, the series behaves well and finds a sum.
- The radius of convergence, therefore, is \( R = 1 \).
- This means, for \( |x| < 1 \), the series will converge, allowing us to calculate a meaningful sum.
Closed Form of Series
A closed form of a series is a neat single expression that represents the sum of an entire series. It’s like compressing a dictionary into a single brief summary. Going back to our original series, \( \sum_{n=1}^{\infty} n x^{2n+1} \), we transformed it into a more manageable form by using the sum of the geometric series we calculated earlier, \( \sum_{n=1}^{\infty} n (x^2)^n = \frac{x^2}{(1-x^2)^2} \).
- Once simplified, the series takes the form: \( x^3 \sum_{n=1}^{\infty} n (x^2)^n \).
- Substituting from the geometric series yields: \( x^3 \cdot \frac{x^2}{(1-x^2)^2} = \frac{x^5}{(1-x^2)^2} \).
Convergence
Convergence is a critical concept when dealing with series. It's like knowing whether a marathon will finish or go on forever. For the series \( \sum_{n=1}^{\infty} n x^{2n+1} \), convergence occurs under specific conditions on \( x \). By rethinking it using only the powers in \( x^2 \), you have it converge as long as the series follows the rule \(|x| < 1\).
- The condition for convergence is derived from the geometric nature of \( \sum_{n=1}^{\infty} n (x^2)^n \).
- Evaluating this, we affirm that the series finds a sum when it's within the "safe zone" defined by the radius of convergence.