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Chapter 9: Problem 46

(Grouping of Terms) Let \(m_{0}:=0\) and let \(m_{1}1} a_{k}\) is convergent, then show that the series $\sum_{k \geq 1} b_{k}$ is convergent and has the same sum. Give an example to show that \(\sum_{k>1} b_{k}\) may be convergent although \(\sum_{k>1} a_{k}\) is divergent.

Short Answer

Expert verified
In this exercise, we showed that if a series \(\sum_{k>1} a_{k}\) converges, then the related series \(\sum_{k \geq 1} b_{k}\) defined by \(b_{k}:=a_{m_{k-1}+1}+\cdots+a_{m_{k}}\) also converges and has the same sum. This is because the series of \(b_{k}\) is a grouping of the terms in the series of \(a_{k}\) based on the given natural numbers \(m_{k}\). As an example demonstrating that the converse is not always true, we considered the harmonic series, which is known to be divergent, and constructed a related convergent \(b_{k}\) series using a specific sequence of natural numbers \(m_{k}\) as described in the step-by-step solution. In this case, the \(b_{k}\) series converged even though the original \(a_{k}\) series (harmonic series) diverged.

Step by step solution

01

Connect the convergence of the two series

Since the series \(\sum_{k>1} a_{k}\) is convergent by assumption, we can write: \[ \lim_{n \to \infty} \left( \sum_{k=1}^{n} a_{k} \right) = L \] where L is the limit or the sum of the series \(\sum_{k>1} a_{k}\). Now let's relate this convergence to \(\sum_{k \geq 1} b_{k}\): \[ \lim_{n \to \infty} \left( \sum_{k=1}^{n} b_{k} \right) = \lim_{n \to \infty} \left( \sum_{k=1}^{n} \left(a_{m_{k-1}+1}+\cdots+a_{m_{k}}\right) \right) \] Notice that the series of \(b_{k}\) is a grouping of the series of \(a_{k}\) based on the given natural numbers \(m_{k}\). Since all terms of \(a_{k}\) are being grouped into the \(b_{k}\) terms, the limit/sum of the \(a_{k}\) series should be equal to the limit/sum of the \(b_{k}\) series.
02

Confirming the convergence of both series

Using the connection between the two series, we can now show the convergence of the \(b_{k}\) series: \[ \lim_{n \to \infty} \left( \sum_{k=1}^{n} b_{k} \right) = \lim_{n \to \infty} \left( \sum_{k=1}^{n} \left(a_{m_{k-1}+1}+\cdots+a_{m_{k}}\right) \right) = L \] This confirms that the series \(\sum_{k \geq 1} b_{k}\) is convergent and has the same sum as the series \(\sum_{k>1} a_{k}\).
03

Provide an example where \(\sum_{k>1} b_{k}\) converges but \(\sum_{k>1} a_{k}\) diverges

Consider the harmonic series, which is known to be divergent: \[ \sum_{k>1} \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ... \] Define the sequence of natural numbers \(m_{k}\) as follows: \[ m_{k} = 2^{k - 1}, \quad k \geq 1 \] such that \(m_{1}=1, m_{2}=2, m_{3} = 4,\cdots\). Now, define the \(b_{k}\) series: \[ b_{k}=\sum_{i=m_{k-1}+1}^{m_{k}} \frac{1}{i}, \quad k \geq 1 \] The \(b_{k}\) series becomes: \[ b_{1} = \frac{1}{1} = 1 \\ b_{2} = \frac{1}{2} \\ b_{3} = \frac{1}{3} + \frac{1}{4} \\ b_{4} = \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \\ \cdots \] Now, examining the sum of the \(b_{k}\) series: \[ \sum_{k>1} b_{k} = 1 + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + ... \] Since each term in parentheses is greater than or equal to \(\frac{1}{2}\), this series is a sum of a geometric series with common ratio of \(1/2\) and the first term also being \(\frac{1}{2}\). Therefore the series converges. So, we have shown how the \(b_{k}\) series can be convergent while the \(a_{k}\) series diverges in this example.

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Most popular questions from this chapter

Let \(f:[1, \infty) \rightarrow \mathbb{R}\) be defined as follows. If \(t \in[1, \infty)\) and \(k \leq t

Let \(a \in \mathbb{R}\) and \(f:[a, \infty) \rightarrow \mathbb{R}\) be such that \(f\) is integrable on \([a, x]\) for all \(x \geq a\). Prove the following: (i) If \(\int_{t \geq a} f(t) d t\) is convergent and \(f(x) \rightarrow \ell\) as \(x \rightarrow \infty\), then \(\ell=0\). (ii) If \(f\) is differentiable and \(\int_{t \geq a} f^{\prime}(t) d t\) is convergent, then there is \(\ell \in \mathbb{R}\) such that \(f(x) \rightarrow \ell\) as \(x \rightarrow \infty\). (Hint: Use part (ii) of Proposition 6.24.) (iii) If \(f\) is differentiable and both \(\int_{t \geq a} f(t) d t\) and \(\int_{t \geq a} f^{\prime}(t) d t\) are convergent, then \(f(x) \rightarrow 0\) as \(x \rightarrow \infty\). (iv) Deduce that the improper integral \(\int_{t \geq 0} t \sin t^{2} d t\) is divergent.

Let \((p, q) \in(0, \infty) \times(0, \infty)\). Show that $$ \beta(p, q)=\int_{0^{+}}^{\infty} \frac{u^{p-1}}{(1+u)^{p+q}} d u=\int_{0^{+}}^{1} \frac{v^{p-1}+v^{q-1}}{(1+v)^{p+q}} d v $$ (Hint: Substitute \(t=u /(1+u)\) and then \(v=1 / u\).)

Find the radius of convergence of the power series \(\sum_{k \geq 0} c_{k} x^{k}\) if for \(k \in \mathbb{N}\), the coefficient \(c_{k}\) is given as follows: (i) \(k !\), (ii) \(k^{2}\), (iii) \(\frac{k}{k^{2}+1}, \quad\) (iv) \(k e^{-k}\), (v) \(c^{k^{2}}\), where \(c \in \mathbb{R}\), (vi) \(\frac{k^{k}}{k !}, \quad\) (vii) \(\frac{2^{k}}{k^{2}}, \quad\) (viii) \(\left(\begin{array}{c}k+m \\ k\end{array}\right)\), where \(m \in \mathbb{N}\).

Give examples to show that if \(\sum_{k} a_{k}\) and \(\sum_{k} b_{k}\) are convergent series of real numbers, then the series \(\sum_{k} a_{k} b_{k}\) may not be convergent. Also show that if \(\sum_{k} a_{k}=A\) and \(\sum_{k} b_{k}=B\), then \(\sum_{k} a_{k} b_{k}\) may be convergent, but its sum may not be equal to \(A B\).
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