Changed layout for more manageable volumes

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-#LyX 2.4 created this file. For more info see https://www.lyx.org/
-\lyxformat 620
-\begin_document
-\begin_header
-\save_transient_properties true
-\origin unavailable
-\textclass book
-\begin_preamble
-\input defs
-\usepackage{amsmath}
-\end_preamble
-\use_default_options true
-\maintain_unincluded_children no
-\language english
-\language_package default
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-\index Index
-\shortcut idx
-\color #008000
-\end_index
-\secnumdepth 3
-\tocdepth 3
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-\end_header
-
-\begin_body
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc1[01]
-\end_layout
-
-\end_inset
-
-What are 
-\begin_inset Formula $H_{0}$
-\end_inset
-
-,
- 
-\begin_inset Formula $H_{1}$
-\end_inset
-
-,
- and 
-\begin_inset Formula $H_{2}$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $H_{0}=0$
-\end_inset
-
-,
- 
-\begin_inset Formula $H_{1}=1$
-\end_inset
-
-,
- 
-\begin_inset Formula $H_{2}=\frac{3}{2}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc4[10]
-\end_layout
-
-\end_inset
-
-Decide which of the following statements are true for all positive integers 
-\begin_inset Formula $n$
-\end_inset
-
-:
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $H_{n}<\ln n$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $H_{n}>\ln n$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $H_{n}>\ln n+\gamma$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-False,
- since 
-\begin_inset Formula $H_{2}=\frac{3}{2}>1>\ln2$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-True,
- because the next one is true.
-\end_layout
-
-\begin_layout Enumerate
-True.
- If this wasn't true for some number 
-\begin_inset Formula $n$
-\end_inset
-
-,
- then it would be
-\begin_inset Formula 
-\[
-0\geq H_{n}-(\ln n+\gamma)>\frac{1}{2n}-\frac{1}{12n^{2}}+\frac{1}{12n^{4}}-\frac{1}{252n^{6}},
-\]
-
-\end_inset
-
-but 
-\begin_inset Formula $\frac{1}{2n}>\frac{1}{12n^{2}}$
-\end_inset
-
- and 
-\begin_inset Formula $\frac{1}{12n^{4}}>\frac{1}{252n^{6}}$
-\end_inset
-
- for any 
-\begin_inset Formula $n\geq1\#$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc9[M18]
-\end_layout
-
-\end_inset
-
-Theorem A applies only when 
-\begin_inset Formula $x>0$
-\end_inset
-
-;
- what is the value of the sum considered when 
-\begin_inset Formula $x=-1$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Using Equation 1.2.6(18) in 
-\begin_inset Formula $(*)$
-\end_inset
-
- and Exercise 1.2.6–48 in 
-\begin_inset Formula $(**)$
-\end_inset
-
-,
-\begin_inset Formula 
-\begin{multline*}
-\sum_{k=1}^{n}\binom{n}{k}(-1)^{k}H_{k}=\sum_{1\leq j\leq k\leq n}\binom{n}{k}(-1)^{k}\frac{1}{j}=\sum_{j=1}^{n}\frac{1}{j}\sum_{k=j}^{n}\binom{n}{k}(-1)^{k}=\\
-=(-1)^{n}\sum_{j=1}^{n}\frac{1}{j}\sum_{k=0}^{n-j}\binom{n}{k}(-1)^{k}\overset{(*)}{=}(-1)^{\cancel{n}}\sum_{j=1}^{n}\frac{1}{j}(-1)^{\cancel{n}-j}\binom{n-1}{n-j}=\\
-=\sum_{j=0}^{n}\frac{1}{j+1}(-1)^{-j-1}\binom{n-1}{n-j-1}=-\sum_{j=0}^{n-1}\frac{(-1)^{j}}{j+1}\binom{n-1}{j}\overset{(**)}{=}\frac{1}{\binom{n}{n-1}}=-\frac{1}{n}.
-\end{multline*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc11[M21]
-\end_layout
-
-\end_inset
-
-Using summation by parts,
- evaluate
-\begin_inset Formula 
-\[
-\sum_{1<k\leq n}\frac{1}{k(k-1)}H_{k}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We have
-\begin_inset Formula 
-\[
-\frac{1}{k(k-1)}H_{k}=\left(\frac{1}{k-1}-\frac{1}{k}\right)H_{k}=\frac{1}{k-1}\left(H_{k-1}+\frac{1}{k}\right)-\frac{1}{k}H_{k},
-\]
-
-\end_inset
-
-so
-\begin_inset Formula 
-\begin{multline*}
-\sum_{k=2}^{n}\frac{H_{k}}{k(k-1)}=\sum_{k=2}^{n}\left(\frac{1}{k-1}\left(H_{k-1}+\frac{1}{k}\right)-\frac{1}{k}H_{k}\right)=1-\frac{1}{n}H_{n}+\sum_{k=2}^{n}\frac{1}{k(k-1)}=\\
-=1-\frac{H_{n}}{n}+\sum_{k=2}^{n}\left(\frac{1}{k-1}-\frac{1}{k}\right)=1-\frac{H_{n}}{n}+H_{n-1}-H_{n}+1=\\
-=1-\frac{1}{n}H_{n}\cancel{+H_{n}}-\frac{1}{n}\cancel{-H_{n}}+1=2-\frac{1}{n}(1+H_{n}).
-\end{multline*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc12[M10]
-\end_layout
-
-\end_inset
-
-Evaluate 
-\begin_inset Formula $H_{\infty}^{(1000)}$
-\end_inset
-
- correct to at least 100 decimal places.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\[
-H_{\infty}^{(1000)}=\sum_{k\geq1}\frac{1}{k^{1000}}=1\pm0.5\cdot10^{-100},
-\]
-
-\end_inset
-
-since 
-\begin_inset Formula $2^{1000}$
-\end_inset
-
- has way more than 100 digits.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc15[M23]
-\end_layout
-
-\end_inset
-
-Express 
-\begin_inset Formula $\sum_{k=1}^{n}H_{k}^{2}$
-\end_inset
-
- in terms of 
-\begin_inset Formula $n$
-\end_inset
-
- and 
-\begin_inset Formula $H_{n}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\begin{align*}
-\sum_{k=1}^{n}H_{k}^{2} & =\sum_{1\leq j\leq k\leq n}\frac{1}{j}H_{k}=\sum_{j=1}^{n}\frac{1}{j}\sum_{k=j}^{n}H_{k}=\sum_{j=1}^{n}\frac{1}{j}\left(\sum_{k=1}^{n}H_{k}-\sum_{k=1}^{j-1}H_{k}\right)\\
- & =\sum_{j=1}^{n}\frac{1}{j}\left((n+1)H_{n}-n-jH_{j-1}+j-1\right)\\
- & =\sum_{j=1}^{n}\left(\frac{1}{j}\left((n+1)H_{n}-n-1\right)-H_{j-1}+1\right)\\
- & =\left((n+1)H_{n}-n-1\right)H_{n}-(nH_{n-1}-n+1)+n\\
- & =(n+1)H_{n}^{2}-(n+1)H_{n}-nH_{n}+n+n\\
- & =(n+1)H_{n}^{2}-(2n+1)H_{n}+2n.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc23[HM20]
-\end_layout
-
-\end_inset
-
-By considering the function 
-\begin_inset Formula $\Gamma'(x)/\Gamma(x)$
-\end_inset
-
-,
- generalize 
-\begin_inset Formula $H_{n}$
-\end_inset
-
- to noninteger values of 
-\begin_inset Formula $n$
-\end_inset
-
-.
- You may use the fact that 
-\begin_inset Formula $\Gamma'(1)=-\gamma$
-\end_inset
-
-,
- anticipating the next exercise.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We have
-\begin_inset Formula 
-\begin{multline*}
-\Gamma'(x)=\lim_{m}\left(\frac{\ln m\cdot m^{x}m!}{x(x+1)\cdots(x+m)}-\frac{m^{x}m!\sum_{k=0}^{m}\frac{x(x+1)\cdots(x+m)}{(x+k)}}{(x(x+1)\cdots(x+m))^{2}}\right)=\\
-=\Gamma(x)\lim_{m}\left(\ln m-\sum_{k=0}^{m}\frac{1}{x+k}\right)=\Gamma(x)(\ln m-H_{k+m}+H_{k-1}).
-\end{multline*}
-
-\end_inset
-
-The fact given in the exercise tells us that
-\begin_inset Formula 
-\[
--\gamma=\Gamma'(1)=\Gamma(1)\lim_{m}(\ln m-H_{m+1}),
-\]
-
-\end_inset
-
-so for 
-\begin_inset Formula $n\in\mathbb{Z}^{>0}$
-\end_inset
-
-,
-\begin_inset Formula 
-\[
-\frac{\Gamma'(n)}{\Gamma(n)}=\lim_{m}(\ln m-H_{n+m}+H_{n-1})=H_{n-1}-\gamma,
-\]
-
-\end_inset
-
-and we can define
-\begin_inset Formula 
-\[
-H_{x}\coloneqq\frac{\Gamma'(x+1)}{\Gamma(x+1)}+\gamma
-\]
-
-\end_inset
-
-for any 
-\begin_inset Formula $x\in\mathbb{C}$
-\end_inset
-
- where this expression is defined or can be extended by continuity.
-\end_layout
-
-\end_body
-\end_document