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\title{Math 10860, Honors Calculus 2}

\author{Homework 9~~~~~~~~~~~~~~~~{\bf NAME}:}

\date{Due by 11pm Tuesday April 14}

\maketitle

\subsection*{Instructions}

Yadda yadda yadda. Email to {\tt math10860homework at gmail.com}.

\subsection*{Reading}          
 
Class notes, Sections 14.4 (on Taylor's theorem) and Chapter 15 (on sequences). 
  
\subsection*{Assignment}   

\begin{enumerate}

%=======
\item 
\begin{enumerate}
\item 
Suppose that $f$ is a function that is defined, and differentiable arbitrarily many times, on some interval $I$ (finite, infinite, whatever). Let $a$ and $x$ be any two numbers in $I$. Suppose that there is some number $M>0$ such that $|f^{(n)}(\xi)| \leq M^n$ for every $n \geq 0$ and every $\xi \in [a, x]$. Prove that 
$$
P_{n,a,f}(x) \rightarrow f(x)~~~\mbox{as $n \rightarrow \infty$}.
$$  
\item 
We saw (in both notes and lectures) that for every $x \in {\mathbb R}$, 
$$
P_{n,0,\exp}(x) \rightarrow \exp(x)~~~\mbox{as $n \rightarrow \infty$}.
$$ 
Generalize this: for every base $b>0$, every $a \in {\mathbb R}$, and every $x \in {\mathbb R}$, the Taylor polynomials of $f(x)=b^x$ centered at $a$ converge to $f(x)$ as $n$ grows. That is,
$$
P_{n,a,b^\cdot}(x) \rightarrow b^x~~~\mbox{as $n \rightarrow \infty$}.
$$
\end{enumerate}
%=======

%=======
\item
\begin{enumerate}
\item Use Theorem 15.2 of the course notes (connecting continuity and limits of sequences) to find, for each fixed $a>0$, $\lim_{n \rightarrow \infty} a^{1/n}$. 
\item Prove a ``squeeze theorem'' for sequences:
\begin{quote}
Let $(a_n), (b_n)$ and $(c_n)$ be sequences with $(a_n), (c_n) \rightarrow L$. If eventually (for all $n > n_0$, for some finite $n_0$) we have $a_n \leq b_n \leq c_n$, then $(b_n) \rightarrow L$ also.
\end{quote}
\item Use the results of parts (a) and (b) to compute
$$
\lim_{n \rightarrow \infty} \left(\frac{2n^2-1}{3n^2+n+2}\right)^\frac{1}{n}.
$$
\end{enumerate}
%=======

%======================
\item Find the following limits:
\begin{enumerate}
\item $\lim_{n \rightarrow \infty} \frac{n}{n+1}$. (For this one, you {\em must} use the definition of sequence limit).

\item $\lim_{n \rightarrow \infty} \sqrt[n]{n^2+n}$. (For this and the remaining parts, a soft argument is fine, meaning, you may freely use theorems proven in lectures and/or notes).


\item $\lim_{n \rightarrow \infty} \left(\sqrt[8]{n^2+1} - \sqrt[4]{n+1}\right)$.


\item $\lim_{n \rightarrow \infty} \left(\frac{n}{n+1} - \frac{n+1}{n}\right)$.


\item $\lim_{n \rightarrow \infty} \frac{2^{n^2}}{n!}$.


\item $\lim_{n \rightarrow \infty} \frac{(-1)^n\sqrt{n}\sin(n^n)}{n+1}$.

\end{enumerate}
%======================


%======================
\item A {\em subsequence} of a sequence
$$
(a_1, a_2, a_3, \ldots)
$$
is a sequence of the form 
$$
(a_{n_1}, a_{n_2}, a_{n_3}, \ldots)
$$
with $n_1 < n_2 < n_3\cdots$. In other words, it is a sequence obtained from another sequence by extracting an infinite subset of the elements of the original sequence, {\em keeping the elements in the same order as they were in the original sequence}.
\begin{enumerate}
\item 
Consider the sequence
$$
\left(\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \cdots\right).
$$
For which numbers $\alpha$ is there a subsequence converging to $\alpha$? 
\item
Now consider the same sequence as in part (a), except remove all duplicated terms, so that it begins
$$
\left(\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \frac{5}{6}, \frac{1}{7}, \cdots\right).
$$
Now for which numbers $\alpha$ is there a subsequence converging to $\alpha$?
\end{enumerate}
%======================

%======================
\item
\begin{enumerate}
\item Prove that if $0 < a < 2$ then $a < \sqrt{2a} < 2$.
\item Prove that the sequence 
$$
\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}, \ldots
$$
converges.
\item Let $a_n$ be the $n$th term of the above sequence, and let $\ell=\lim_{n \rightarrow \infty} a_n$. Carefully applying a theorem proved in lectures, find $\ell$.
\end{enumerate}
%======================

%======================
\item This question provides a useful estimate on $n!$: $n! \approx (n/e)^n$.
\begin{enumerate}
\item Show that if $f:[1,\infty)$ is increasing then
$$
f(1)+\cdots + f(n-1) < \int_1^n f(x)dx < f(2) + \cdots + f(n).
$$
\item By taking $f=\log$ deduce that
$$
\frac{n^n}{e^{n-1}} < n! < \frac{(n+1)^{n+1}}{e^n}.
$$
\item Deduce that\footnote{Note that this only says that for large $n$, $\sqrt[n]{n!}$ is close to $n/e$; it does not say that for large $n$, $n!$ is close to $(n/e)^n$ --- it is not. In fact, all we can get out of the bounds in part b) is that
$$
e\left(\frac{n}{e}\right)^n < n! < e(n+1)\left(\frac{n}{e}\right)^n.
$$
A better, and much more difficult to prove, bound on $n!$ is given by {\em Stirling's formula}:
$$
\lim_{n \rightarrow \infty} \frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n} = 1;
$$
in other words, for all $\varepsilon > 0$ there is $n_0$ such that $n > n_0$ implies
$$
(1-\varepsilon)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n < n! < (1+\varepsilon)\sqrt{2\pi n}\left(\frac{n}{e}\right)^n.
$$}
$$
\lim_{n \rightarrow \infty} \frac{\sqrt[n]{n!}}{n} = \frac{1}{e}.
$$
\end{enumerate}
%======================



%======================
\item The {\em Harmonic number} $H_n$ is the number $H_n = \sum_{k=1}^n \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$. This exercise gives a very useful estimate on $H_n$, namely $H_n \approx \log n$.
\begin{enumerate}
\item Notice that $H_1 = 1$, $H_2 = 1 + \frac{1}{2}$ and 
$$
H_4 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \geq 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1 + \frac{2}{2}.
$$
Generalize this: prove that for all $k \geq 0$, $H_{2^k} \geq 1+\frac{k}{2}$ (and so $(H_n)_{n=1}^\infty$ diverges to $+\infty$). 
\item Prove that for all natural numbers $n$,
$$
\frac{1}{n+1} < \log(n+1) - \log n < \frac{1}{n}.
$$
\item Deduce from part (b) that the sequence $(H_n-\log n)_{n=2}^\infty$ is decreasing and bounded below by $0$.
\item Explain why you can deduce that there is a number $\gamma \geq 0$ such that 
$$
\lim_{n \rightarrow \infty} \left(H_n - \log n\right) = \gamma.
$$
(This number is known as the {\em Euler-Mascheroni constant}, and is approximately $0.57721$. It is not known whether $\gamma$ is rational or irrational.)
\end{enumerate}
%======================

\end{enumerate} 

\end{document}

