Problem 4 Show that if $X$ and $Y$ are... [FREE SOLUTION]

91影视

Introduction to Real Analysis

Robert G. Bartle, Donald R. Sherbert

$Math Studyset 91影视 Explanations$ Math

3 Edition

Chapter 3: Problem 4

Show that if $X$ and $Y$ are sequences such that $X$ converges to $x \neq 0$ and $X Y$ converges, then $Y$ converges.

Short Answer

Expert verified

The conclusion is that if $X$ and $Y$ are sequences such that $X$ converges to $x \neq 0$ and $XY$ converges, then $Y$ also converges.

Step by step solution

Understand sequence convergence

Understanding sequence convergence is key to solving this problem. A sequence converges towards a particular value if, for every positive real number $ \epsilon $, there exists an $ N $, such that for all values $ n > N $, the absolute difference $ |a_n - a| < \epsilon $ holds true, where $ a_n $ is an element of the sequence and $ a $ is the value the sequence converges to.

Consider sequences $X$ and $XY$

We have two sequences, $ X, XY $ both converging. It is given that $ X $ converges to $ x \neq 0 $, which means for all $ \epsilon > 0 $, there exists an $ N1 $ such that $ |X_n - x| < \epsilon $, for all $ n > N1 $. Similarly, $ XY $ converges to a certain value, which we do not know. This means that for all $ \epsilon > 0 $, there exists an $ N2 $, such that $ |(XY)_n - xy| < \epsilon $, for all $ n > N2 $

Prove sequence $ Y $ convergence

Now we have to prove that $Y$ converges, for that, let $z$ be the limit point of $Y$. That means for all $ \epsilon > 0 $, there exists an $ N $ such that $ |Y_n - z| < \epsilon $, for all $ n > N $. We can express $Y_n$ as $ \frac{(XY)_n}{X_n} $. This implies $ |Y_n - z| = | \frac{(XY)_n - xz}{X_n} | $. But, we know that the numerator in this equation converges as $n$ tends towards infinity and since $X_n$ is not zero, it implies $Y_n$ converges to $z$. Therefore, we can conclude that $Y$ converges.

Conclusion

From the above steps, we can verify that if $X$ and $Y$ are sequences such that $X$ converges to $x \neq 0$ and $XY$ converges, then $Y$ also converges.

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91影视

Show that if \(X\) and \(Y\) are sequences such that \(X\) converges to \(x \neq 0\) and \(X Y\) converges, then \(Y\) converges.

Short Answer

Step by step solution

Understand sequence convergence

Consider sequences \(X\) and \(XY\)

Prove sequence \( Y \) convergence

Conclusion

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