#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
\inputencoding utf8
\fontencoding auto
\font_roman "default" "default"
\font_sans "default" "default"
\font_typewriter "default" "default"
\font_math "auto" "auto"
\font_default_family default
\use_non_tex_fonts false
\font_sc false
\font_roman_osf false
\font_sans_osf false
\font_typewriter_osf false
\font_sf_scale 100 100
\font_tt_scale 100 100
\use_microtype false
\use_dash_ligatures true
\graphics default
\default_output_format default
\output_sync 0
\bibtex_command default
\index_command default
\float_placement class
\float_alignment class
\paperfontsize default
\spacing single
\use_hyperref false
\papersize default
\use_geometry false
\use_package amsmath 1
\use_package amssymb 1
\use_package cancel 1
\use_package esint 1
\use_package mathdots 1
\use_package mathtools 1
\use_package mhchem 1
\use_package stackrel 1
\use_package stmaryrd 1
\use_package undertilde 1
\cite_engine basic
\cite_engine_type default
\biblio_style plain
\use_bibtopic false
\use_indices false
\paperorientation portrait
\suppress_date false
\justification true
\use_refstyle 1
\use_formatted_ref 0
\use_minted 0
\use_lineno 0
\index Index
\shortcut idx
\color #008000
\end_index
\secnumdepth 3
\tocdepth 3
\paragraph_separation indent
\paragraph_indentation default
\is_math_indent 0
\math_numbering_side default
\quotes_style english
\dynamic_quotes 0
\papercolumns 1
\papersides 1
\paperpagestyle default
\tablestyle default
\tracking_changes false
\output_changes false
\change_bars false
\postpone_fragile_content false
\html_math_output 0
\html_css_as_file 0
\html_be_strict false
\docbook_table_output 0
\docbook_mathml_prefix 1
\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