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Chapter 6: Problem 5

Suppose the power series \(\sum_{n=0}^{\infty} a_{n} s^{n}\) is convergent for \(|s|

Short Answer

Expert verified
Question: Prove that if a power series \(\sum_{n=0}^{\infty} a_{n} s^{n}\) converges for \(|s|<R\), then its derived series \(\sum_{n=1}^{\infty} n a_{n} s^{n-1}\) is also convergent for \(|s|<R\). Answer: By applying the Ratio Test to both the original series and the derived series, and showing that the limit of the derived series is less than 1 for the same interval \(|s|<R\), we can conclude that \(\sum_{n=1}^{\infty} n a_{n} s^{n-1}\) is convergent for \(|s|<R\).

Step by step solution

01

Use the Ratio Test

Apply the Ratio Test for convergence to the power series \(\sum_{n=0}^{\infty} a_{n} s^{n}\). The Ratio Test states that the series converges if the limit as \(n \to \infty\) of the absolute value of the ratio of consecutive terms is less than 1. This means we need to find: \(\lim_{n\to\infty}\left|\frac{a_{n+1} s^{n+1}}{a_{n} s^{n}}\right| = \lim_{n\to\infty}\left|s\frac{a_{n+1}}{a_{n}}\right|\) Given that the seris converges, this limit must be less than 1 for \(|s|<R\).
02

Apply the Ratio Test to the derived series

Now we will apply the Ratio Test to the derived series \(\sum_{n=1}^{\infty} n a_{n} s^{n-1}\). We are considering the limit: \(\lim_{n\to\infty}\left|\frac{(n+1) a_{n+1} s^{n}}{n a_{n} s^{n-1}}\right| = \lim_{n\to\infty}\left|\frac{(n+1) a_{n+1}}{n a_{n}} s\right|\) Notice that the term without \(s\) is \(\frac{(n+1) a_{n+1}}{n a_{n}}\) which is very similar to the term from the initial series.
03

Compare the limits

Now we compare the limits obtained for the original series and the derived series. The derived series limit is just the original series limit times \(s\): \(\lim_{n\to\infty}\left|\frac{(n+1) a_{n+1}}{n a_{n}} s\right| = \lim_{n\to\infty} \left|s\frac{a_{n+1}}{a_{n}}\right| |s| = \left(\lim_{n\to\infty} \left|s\frac{a_{n+1}}{a_{n}}\right|\right) |s|\) Since the initial series converges for \(|s|<R\), we know that \(\lim_{n\to\infty} \left|s\frac{a_{n+1}}{a_{n}}\right|\) is less than 1. Therefore, the derived series limit is the converging series limit times \(|s|\), which will also be less than 1 for \(|s|<R\).
04

Final conclusion

Since the limit for the derived series is less than 1 in the same interval \(|s|<R\), by the Ratio Test, we can conclude that \(\sum_{n=1}^{\infty} n a_{n} s^{n-1}\) is convergent for \(|s|<R\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ratio Test
The Ratio Test is a crucial tool for determining the convergence of infinite series, especially power series. When applied, it helps us understand if a series will converge, which means to have a finite sum.
To apply the Ratio Test, we examine the limit of the absolute value of the ratio of consecutive terms of a series as the variable approaches infinity. Specifically, for a series \(\sum a_n\), you look at:
  • \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\)
If this limit is less than 1, the series converges absolutely. If it's greater than 1, or is infinite, the series diverges. If it equals 1, the test is inconclusive.
This test is particularly useful because of how it handles series with factorials or exponentials. In our exercise, the Ratio Test confirms that the original power series \(\sum a_n s^n\) converges for \(|s| < R\). Similarly, we apply it to the derived series to confirm it also converges for the same interval \(|s| < R\).
Interval of Convergence
The interval of convergence refers to all values of the variable for which a power series converges. Determining this interval is essential for understanding the behavior and limits of the series.
For a power series \(\sum a_n s^n\), there is usually a specific radius, \( R \), such that:
  • The series converges absolutely for \(|s| < R\).
  • The series diverges for \(|s| > R\).
  • The behavior at \(|s| = R\) generally requires individual testing.
In our context, the original series converges within \(|s| < R\). By showing that the derived series also converges within this same interval using the Ratio Test means we don't need to extend or shrink this interval. The confirmed interval ensures that both series behave predictably within these bounds, providing consistency in calculations and applications.
Derived Series
Derived series often come up when dealing with derivatives of power series, hence the name. In this context, we're referring to a new series derived from the original by a transformation.
In the exercise, we examined the series \(\sum na_n s^{n-1}\), derived from \(\sum a_n s^n\). The elements were rearranged and multiplied by \(n\), changing the focus to the convergence of this new form.
This process involves:
  • Adjusting terms by a factor, such as \(n\), that changes the series indexation or term power.
  • Checking the convergence of this modified series, especially within known intervals.
Applying the Ratio Test to this derived series showed it still converged under the same conditions as the original. Such manipulations frequently appear in calculus and analysis, offering powerful techniques for differential equations and functional analysis. Understanding derived series provides a deeper insight into the behavior and flexibility of series across mathematical applications.

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Most popular questions from this chapter

Using only the definition of compactness, show that: (i) \((a, b)\) is not a compact subset of \(\mathbb{R}\) (ii) \(\left\\{(x, y): x^{2}+y^{2}<1\right\\}\) is not a compact subset of \(\mathbb{R}^{2}\).

Define a sequence of polynomials inductively on \([-1,1]\) by setting \(p_{0}(t)=1\) and $$ p_{n+1}=p_{n}(t)+\frac{1}{2}\left(t^{2}-p_{n}^{2}(t)\right), \quad n=0,1, \ldots $$ Show that \(0 \leq p_{n}(t) \leq|t|\) for \(-1 \leq t \leq 1\) and hence that \(p_{n}(t) \rightarrow|t|\) uniformly on \([-1,1]\).

The following construction shows how to prove directly that \([0,1]\) is a compact subset of \(\mathbb{R}\). Let \(\mathcal{G}=\left\\{G_{\alpha}\right\\}\) be a collection of open sets in \(\mathbb{R}\) which cover \([0,1]\). Then 0 belongs to some \(G_{\alpha}\). Since this set \(G_{\alpha}\) is open, \([0, \epsilon) \subset G_{\alpha}\) for some \(\epsilon>0\). Set \(x=\sup \\{y \in[0,1]:[0, y]\) is contained in the union of a finite number of members of \(\mathcal{G}\\}\). Show that in fact \(x=1\).

Let \(f\) be a uniformly continuous mapping from one normed linear space \(X\) into another, \(Y\). Show that the image of a totally bounded set in \(X\) is a totally bounded set in \(Y\). Is this true if \(f\) is only assumed to be continuous?

Let $$ f_{n}(s)=\frac{n s}{n^{2}+s^{2}}, \quad n \geq 1 $$ Show that \(f_{n}(s) \rightarrow 0\) pointwise for all \(s>0\). Does \(f_{n} \rightarrow 0\) uniformly (i) on \([0, \infty) ?\) (ii) on \([0,1] ?\)
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