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Chapter 3: Problem 2

Give an example of two divergent sequences \(X\) and \(Y\) such that: (a) their sum \(X+Y\) converges, (b) their product \(X Y\) converges.

Short Answer

Expert verified
So, sequences \(X\) and \(Y\), with \(X_n = -1^n\) and \(Y_n = -1^{n+1}\), are two examples of divergent sequences wherein their sum \(X+Y\) converges to 0 and their product \(X Y\) converges to 1.

Step by step solution

01

Defining Sequences X and Y

Choose two sequences. For sequence \(X\): let the nth term \(X_n = -1^n = -1, 1, -1, 1, -1, ...\)For sequence \(Y\): let the nth term \(Y_n = -1^{n+1} = 1, -1, 1, -1, 1, ...\)These sequences \(X\) and \(Y\) are divergent as they do not approach a specific limit but continue to oscillate between two values.
02

Verify the Convergence of Sum \(X + Y\)

Let's add the terms of sequences \(X\) and \(Y\) to examine if their sum \(X + Y\) converges.The nth term of \(X + Y\) is given by \(X_n + Y_n = -1^n + -1^{n+1}\).Regardless of the value of \(n\), the sum is always zero, i.e., \(X + Y = 0, 0, 0, 0, ...\), which is clearly a convergent sequence. So, the sum of sequences \(X\) and \(Y\) converges to 0.
03

Verify the Convergence of Product \(X Y\)

Now let's calculate the product of the terms of sequences \(X\) and \(Y\) to check if their product \(X Y\) converges.The nth term of \(X Y\) is given by \(X_n \times Y_n = (-1^n) \times (-1^{n+1})\).Regardless of the value of \(n\), the product is always 1, i.e., \(X Y = 1, 1, 1, 1, ...\), which is clearly a convergent sequence. So, the product of sequences \(X\) and \(Y\) converges to 1.

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Most popular questions from this chapter

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, and if \(b_{n}:=\left(a_{1}+\cdots+a_{n}\right) / n\) for \(n \in \mathbb{N}\), then show that \(\sum b_{n}\) is always divergent.

Suppose that \(x_{n} \geq 0\) for all \(n \in \mathbb{N}\) and that \(\lim \left((-1)^{n} x_{n}\right)\) exists. Show that \(\left(x_{n}\right)\) converges.

Show that \(\lim \left(2^{n} / n !\right)=0 .\) [Hint: if \(n \geq 3\), then \(0<2^{n} / n ! \leq 2\left(\frac{2}{3}\right)^{n-2}\).]

Use the Cauchy Condensation Test to discuss the \(p\) -series \(\sum_{n=1}^{\infty}\left(1 / n^{p}\right)\) for \(p>0\).

Show that the following sequences are divergent. (a) \(\left(1-(-1)^{n}+1 / n\right)\), (b) \((\sin n \pi / 4)\).
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