Math 521 midterm 2

The second midterm will cover the material in chapter 3 of Rudin’s book which deals with sequences and series.

Review questions

  1. Let $(X,d)$ be a metric space, and let $x_n\in X$ be a sequence. State the definition of “$x_n\to p$”.
  2. Show that a sequence $x_n\in X$ cannot have more than one limit.
  3. State the definition of a Cauchy sequence.
  4. State the definition of “complete metric space”
  5. Is a compact metric space complete?
  6. Is a complete metric space compact?
  7. Show that a bounded and monotone sequence of real numbers converges.
  8. State the definition of “lim sup” of a sequence.
  9. State the definition of convergence of a series $\sum_{n=0}^\infty a_n$ whose terms are real numbers.
  10. Show that a series with positive terms converges if and only if its partial sums are bounded.
  11. Show that the geometric series $\sum_{n=0}^\infty x^n$ converges if and only if $|x|\lt 1$.
  12. Show that the series $\sum_{n=1}^\infty \frac1n$ diverges without using the “condensation test”—alternatively, prove the condensation test.
  13. Show that the series $\sum_1^\infty \frac{1}{n^2}$ converges without using the “condensation test”—alternatively, prove the condensation test.
  14. State and prove the “comparison test” (theorem 3.25 in Rudin). What does the proof have to do with Cauchy sequences?
  15. State the definition of absolute convergence for series.
  16. Give an example of a convergent series which is not absolutely convergent.
  17. State and prove the alternating series test.

Practice problems

These are mostly the homework problems that were assigned.
  1. Consider the sequence of real numbers given by $a_n=n^{-1}$.
    1. Show that $a_n\to 0$, directly from the definition.
      Given $\varepsilon\gt0$ choose an integer $N$ such that $N\gt 1/\varepsilon$. Then for all $n\ge N$ one has $0\lt a_n \le 1/N \lt \varepsilon$.
    2. Show that if a sequence $x_n$ converges, then any subsequence of $x_n$ also converges and has the same limit.
      Let $y_k = x_{n_k}$ be a subsequence of $x_n$, and let $L$ be the limit of $x_n$. In particular this implies that $n_k\ge k$ for all $k$. Given $\varepsilon\gt0$ we choose $N\in\N$ such that $|x_n-L|\lt \varepsilon$ for all $n\ge N$. Then, if $k\ge N$ we have $n_k\ge k\ge N$ and thus $|y_k-L| = |x_{n_k}-L| \lt \varepsilon$.
    3. Show that $b_n=n^{-2}\to 0$ without using the definition of convergence.
      $b_n = a_{n^2}$, so $b_n$ is a subsequence of the sequence $a_n=1/n$. Therefore it converges, and it has the same limit: $\lim_{n\to\infty} 1/n^2 = 0$.
  2. Let $x_n$ be a convergent sequence of points in a metric space $(X,d)$. Show that the sequence is bounded.
    Let $x_n\to p$, i.e. let $p$ be the limit of the sequence $x_n$. Choose $\varepsilon=1$. There is an $N\in\N$ such that for all $n\ge N$ on has $d(x_n, p)\lt \varepsilon=1$. Thus, if \[ R=\max\left\{1, d(x_1, p), \dots, d(x_{N-1}, p)\right\}, \] then all $x_k$ satisfy $d(x_k, p)\le R$.
  3. Let $x_n$ be a sequence of points in a metric space $(X,d)$. Show that $x_n\to x$ holds if and only if $d(x_n, x)\to0$.
    Consider the sequence of real numbers $a_n = d(x_n, p)$. We are asked to show that $a_n\to0 \iff x_n\to p$.

    Assume $a_n\to 0$. To show $x_n\to p$, let $\varepsilon\gt 0$ be given. By assumption there is an $N\in\N$ such that $|a_n|\lt \varepsilon$ for all $n\ge N$. This implies $d(x_n, p)\lt \varepsilon$ for all $n\ge N$. Therefore $x_n\to p$.

    Assume $x_n\to p$. To show $a_n\to 0$, let $\varepsilon\gt 0$ be given. By assumption there is an $N\in\N$ such that $d(x_n, p)\lt \varepsilon$ for all $n\ge N$. This implies $|a_n|\lt \varepsilon$ for all $n\ge N$. Therefore $a_n\to 0$.

  4. Let $(X,d)$ be a metric space, $C\subset X$ a closed subset, and $x_n\in C$ a sequence of points with $x_n\to x$. Show that $x\in C$.
    In order to reach a contradiction, assume that $x\not\in C$. The complement $C^c$ of $C$ is open in $X$, so there is an $\varepsilon\gt0$ such that $B_\varepsilon(x)\subset C^c$. Since $x_n\to x$, there is an $N_\varepsilon\in\N$ for such that $x_n\in B_\varepsilon(x)$ for all $n\ge N$. But then we would have $x_n\not\in C$ for $n\ge N$, which is a contradiction. We conclude that $x\in C$ after all.
  5. Let $x_n\in\R^k$ and $y_n\in \R^k$ be two sequences of vectors for which $x_n\to x$ and $y_n\to y$. Show that $(x_n, y_n) \to (x,y)$. Here $(x,y)$ is the dot product, or inner product of the two vectors. In Rudin’s notation (page 16): the assignment is to show that $x_n\cdot y_n \to x\cdot y$.
    The sequences $x_n\in\R^k$ and $y_n\in\R^k$ converge, so they are bounded, i.e. there is an $R\gt0$ such that $\|x_n\|\le R$, and $\|y_n\|\le R$ for all $n\in\N$. We also have \begin{align*} \left|x_n\cdot y_n - x\cdot y\right| &=\left| x_n\cdot y_n - x_n\cdot y + x_n\cdot y - x\cdot y\right| \\ &\le \left| x_n\cdot y_n - x_n\cdot y\right| + \left|x_n\cdot y - x\cdot y\right| & \text{triangle }\le\\ &\le \left| x_n\cdot (y_n - y)\right| + \left|(x_n - x)\cdot y\right| \\ &\le \| x_n\|\,\| y_n - y\| + \|x_n - x\|\,\| y\| & \text{Cauchy }\le \\ &\le R\| y_n - y\| + R \|x_n - x\| \end{align*} Since $x_n$ and $y_n$ converge we have \[ \lim_{n\to\infty} R\| y_n - y\| + R \|x_n - x\| = 0 \] and hence \[ \lim_{n\to\infty} \left|x_n\cdot y_n - x\cdot y\right| =0. \]
  6. Let $A\subset\R$, and assume that every term in the sequence $\{x_n\}_{n\in\N}$ is an upper bound for $A$. Show that if $x_n\to x$, then $x$ is also an upper bound for $A$.
    Suppose $x$ is not an upper bound for $A$. Then there is a number $a\in A$ with $a\gt X$. Let $\varepsilon= a-x$. Since $x_n\to X$, there is an $N$ such that for all $n\ge N$ one has $|x_n-x|\lt \varepsilon$. This implies, for $n\ge N$, that $x_n \lt x+\varepsilon = a$. Hence $x_n$ is not an upper bound for $A$ if $n\ge N$, contradicting what was given.
  7. Instead of the definition in Rudin (3.16, page 56) we will use the following alternative definition of “lim sup”, which was discussed in class: \[ \limsup_{n\to\infty} x_n = \lim_{n\to\infty}\Bigl\{\sup_{k\ge n} x_k\Bigr\}. \] This is currently the more common definition: see, for example, the wikipedia page on lim inf and lim sup.
  8. Prove that if $\limsup_{n\to\infty} a_n =A$ then for every $\varepsilon\gt0$ and any $n\in\N$ there is an $m\gt n$ such that $a_m\gt A-\varepsilon$.
    Let $\varepsilon\gt0$ and $n\in\N$ be given. By definition of “lim sup” we have \[ \lim_{m\to\infty} \Bigl(\sup\bigl\{a_k : k\ge m\bigr\}\Bigr) =A. \] By definition of “limit” this implies that there is an $N_\varepsilon\in\N$ such that for all $m\ge N_\varepsilon$ we have \[ A - \varepsilon \lt \sup\bigl\{a_k : k\ge m\bigr\}\lt A + \varepsilon \] We choose $m \gt \max \{N_\varepsilon, n\}$. Then \[ \sup\{a_k : k\ge m\} \gt A - \varepsilon. \] Since $\sup \{\dots\}$ is the least upper bound it follows that $A- \varepsilon$ is not an upper bound for $\{a_k : k\ge m\}$. Thus there is an $k\ge m$ such that $a_k \gt A-\varepsilon$. We had chosen $m\gt n$, and thus $k\gt n$, so that we have found a $k\gt n$ for which $a_k\gt A- \varepsilon$.
  9. Compute $\limsup_{n\to\infty} a_n$ and $\liminf_{n\to\infty} a_n$ for the following sequences, showing how your result follows from the definition of limsup/inf (given below):
    1. $a_n = (-1)^n$
    2. $a_n = (-1)^n+ \frac 2n$
    3. $a_n = (-1)^n \frac {n+2}n$
    4. $a_n = n$
    5. $a_n = (-1)^n n$
    6. $a_n = \bigl(1 + (-1)^n\bigr) n$
    In each case compute $m_n = \inf \{a_k : k\ge n\}$, and $M_n = \sup \{a_k : k\ge n\}$. Then $\lim_{n\to\infty} m_n$ and $\lim_{n\to\infty}M_n$ are the lim inf and lim sup, respectively.
    1. $a_n = (-1)^n$: $m_n=-1$ and $M_n=+1$ for all $n$, so $\liminf_{n\to\infty} a_n = -1$, while $\limsup_{n\to\infty}a_n = +1$.
    2. $a_n = (-1)^n+ \frac 2n$: $m_n = -1$ for all $n$. $M_{n} = 1+\frac2n$ if $n$ is even, and $M_n=M_{n+1}$ if $n$ is odd. Hence $\liminf_{n\to\infty} a_n = -1$, while $\limsup_{n\to\infty}a_n = +1$.
    3. $a_n = (-1)^n \frac {n+2}n$: $m_n=-\frac{n+2}n$ if $n$ is odd, and $m_{n}=m_{n+1}$ if $n$ is even, so $\liminf_{n\to\infty} a_n = -1$; $M_n=\frac{n+2}n$ if $n$ is even, and $M_{n}=M_{n+1}$ if $n$ is odd, so $\limsup_{n\to\infty} a_n = 1$.
    4. $a_n = n$: $m_n=n$, and $M_n=+\infty$ for all $n$. Both lim sup and lim inf are $+\infty$.
    5. $a_n = (-1)^n n$: $m_n=-\infty$, $M_n=+\infty$. $\liminf_{n\to\infty} a_n = -\infty$, while $\limsup_{n\to\infty}a_n = +\infty$.
    6. $a_n = \bigl(1 + (-1)^n\bigr) n$ : $m_n=0$, $M_n= 2n$ if $n$ is even, $M_n=M_{n+1}$ if $n$ is odd. $\liminf_{n\to\infty} a_n = 0$, while $\limsup_{n\to\infty}a_n =\infty$.
  10. Let $x_n=\cos(\theta + n\frac\pi3)$ where $\theta\in(0, \frac\pi3)$ is some constant. For this sequence, compute
    1. $\sup_{n\in\N} x_n$
    2. all possible limits of subsequences of $x_n$
    3. $\limsup_{n\to\infty} x_n$
    See student-solution on Piazza (type “@52” in the search box at the top left of the Piazza page).
  11. Let $(X,d)$ be a metric space, and let $x_n$ be a convergent sequence in $X$. Show that $x_n$ also is a Cauchy sequence.
    Assume $x_n\to p$. Let $\varepsilon\gt0$ be given. Then there is an $N_\varepsilon\in\N$ such that for all $n\ge N_\varepsilon$ one has $d(x_n, p)\lt \varepsilon$. If $n, m\ge N_\varepsilon$, then \[ d(x_n, x_m) \le d(x_n, p)+d(p, x_m) \le \frac \varepsilon2 + \frac\varepsilon2 =\varepsilon. \] Thus $\{x_n\}$ is a Cauchy sequence.
  12. Give an example of a metric space $(X,d)$ with a Cauchy sequence that does not converge.
    There are many examples. Perhaps the simplest example is to take any metric space $(X,d)$ and a convergent sequence $x_n\in X$, $p=\lim_{n\to\infty}x_n$, and then consider the space $(Y, d)$ where $Y=X\setminus \{p\}$, and $d$ is the same metric as on $X$. For instance, let $Y=\R\setminus\{0\}$, with distance $d(x, y) = |x-y|$; then the sequence $x_n = \frac1n$ is a Cauchy sequence, but it does not converge in $Y$.
  13. Give an example of a complete metric space that is not compact. Provide a brief explanation for your answer.
    The real line. Compact metric spaces are bounded, and while $\R$ is complete, it is not bounded.
  14. Let $x_n$ and $y_n$ be two bounded sequences of real numbers. Show that \[ \limsup_{n\to\infty} (x_n+y_n) \le \limsup_{n\to\infty} x_n + \limsup_{n\to\infty} y_n. \]
    By definition \[ \limsup_{n\to\infty}(x_n+y_n) = \lim_{n\to\infty} \Bigl(\sup\{x_k+y_k : k\ge n\}\Bigr). \] For every $k\ge m$ we have \[ x_k \le \sup\{x_l : l\ge m\}, \qquad y_k \le \sup\{y_l : l\ge m\}. \] Hence, for all $k\ge m$, \[ x_k+y_k \le \sup\{x_l : l\ge m\} + \sup\{y_l : l\ge m\}. \] The quantity on the right does not depend on $k$ and thus it is an upper bound for $\{x_k+y_k : k\ge m\}$. It is therefore not smaller than the least upper bound: \[ \sup\{x_k+y_k : k\ge m\} \le \sup\{x_l : l\ge m\} + \sup\{y_l : l\ge m\}. \] If we now take the limit for $m\to\infty $ on both sides we get \[ \limsup_{k\to\infty} x_k+y_k \le \limsup_{k\to\infty} x_k + \limsup_{k\to\infty} y_k. \]
  15. Find two sequences $x_n$, $y_n$, of real numbers for which \[ \limsup_{n\to\infty} (x_n+y_n) \lt \limsup_{n\to\infty} x_n + \limsup_{n\to\infty} y_n. \]
    There are many possibilities. The following was suggested by a student in my office hour: \begin{align*} \{x_n\} &= \{+1, 0, +1, 0, +1, 0, +1, 0, \dots \}\\ \{y_n\} &= \{-1, 0, -1, 0, -1, 0, -1, 0, \dots \} \end{align*} For these sequences $x_n+y_n = 0$ for all $n$, so $\limsup_{n\to\infty}x_n+y_n =0$. But \[ \limsup x_n = 1, \qquad \limsup y_n =0, \] so $\limsup(x_n+y_n) \lt \limsup x_n+ \limsup y_n$.
  16. If $x_n$ and $y_n$ are bounded sequences of real numbers for which $x_n$ converges, then show that \[ \limsup_{n\to\infty} (x_n+y_n) = \lim_{n\to\infty} x_n + \limsup_{n\to\infty} y_n. \]
    Abbreviate \[ \lim_{n\to\infty} x_n = X, \qquad \limsup_{n\to\infty}y_n = Y. \] By definition \[ \limsup_{n\to\infty} \bigl(x_n +y_n\bigr) = \lim_{n\to\infty} \Bigl\{\sup_{k\ge n} \bigl(x_n+y_n\bigr) \Bigr\}. \] Let $\varepsilon\gt 0$ be given. Then there is an $N_\varepsilon$ such that for all $n\ge N_\varepsilon$ we have \[ X-\varepsilon \lt x_n \lt X+\varepsilon, \] and \[ Y - \varepsilon \lt \sup_{k\ge n} y_k \lt Y + \varepsilon. \] For any $k\ge N_\varepsilon$ we then have \[ X-\varepsilon + y_k \lt x_k + y_k \lt X+\varepsilon +y_k, \] and thus for any $n\ge N_\varepsilon$, \[ X-\varepsilon + \sup_{k\ge n}y_k \lt \sup_{k\ge n} \bigl(x_k + y_k\bigr) \lt X+\varepsilon +\sup_{k\ge n}y_k, \] and therefore, for all $n\ge N_\varepsilon$, \[ X+ Y - 2\varepsilon \lt \sup_{k\ge n} \bigl(x_k + y_k\bigr) \lt X + Y + 2\varepsilon. \] This implies, by definition of “$\lim_{n\to \infty} (\dots)$” that \[ \lim_{n\to\infty} \Bigl\{\sup_{k\ge n} \bigl(x_k + y_k\bigr) \Bigr\} = X+Y. \]