\documentclass[10pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage[version=4]{mhchem}
\usepackage{stmaryrd}
\usepackage{bbold}
\title{HW solutions }
\author{bencheikh mohammed el amine}
\date{}
\begin{document}
\maketitle
\begin{enumerate}
\item (Kummer's Test) Let $\sum_{n \geq 0} u_{n}$ be a positive series, and let $\sum_{n \geq 0}
d_{n}$ be a positive divergent series. Define the Kummer characteristic $K_{n}$ by
\end{enumerate}
$$
K_{n}=\frac{1}{d_{n}} \cdot \frac{u_{n}}{u_{n+1}}-\frac{1}{d_{n+1}}, \quad n \geq 0
$$
Suppose that $\lim _{n \rightarrow \infty} K_{n}=K$ (including the infinite limits $K= \pm \infty$ ).
Show that
(a) If $K>0$, then $\sum_{n>0} u_{n}$ is convergent.
(b) If $K<0$, then $\sum_{n \geq 0} u_{n}$ is divergent.\\
fbox{Proof. We begin by considering the case where $L>0$. Choose a number $r$ such that $0<r<L$.
Then there must exist some integer $N>0$ such that
$$
B_{n} \frac{a_{n}}{a_{n+1}}-B_{n+1}>r \Leftrightarrow B_{n} a_{n}-B_{n+1} a_{n+1}>r a_{n+1}
$$
for all $n \geq N$. Given any positive integer $m$, we have
$$
\begin{aligned}
B_{N} a_{N}-B_{N+1} a_{N+1} & >r a_{N+1} \\
B_{N+1} a_{N+1}-B_{N+2} a_{N+2} & >r a_{N+2} \\
B_{N+2} a_{N+2}-B_{N+3} a_{N+3} & >r a_{N+3} \\
\vdots & \\
B_{N+m-1} a_{N+m-1}-B_{N+m} a_{N+m} & >r a_{N+m}
\end{aligned}
$$
And then by adding these inequalities together, we get cancellations of all of the terms on the left
side of the inequality except for the first and last. Thus we have
$$
B_{N} a_{N}-B_{N+m} a_{N+m}>r\left(a_{N+1}+\cdots+a_{N+m}\right)=r\left(S_{N+m}-S_{N}\right),
$$
where $S_{n}$ is the partial sum for $\sum_{k=N+1}^{N+m} a_{k}$. It follows that
$$
r S_{N+m}<r S_{N}+B_{N} a_{N}-B_{N+m} a_{N+m}<r S_{N}+B_{N} a_{N} .
$$
Let $C$ be the constant $\frac{\left(r S_{N}+B_{N} a_{N}\right)}{r}$. The above inequality tells us that
$S_{n}<C$ for all $n \geq N$. Thus, the sequence of partial sums $\left\{S_{n}\right\}$ of the series $\
sum_{n=1}^{\infty} a_{n}$ is bounded and therefore the series converges.
Now consider the case where $L<0$. This means that there exists an integer $N>0$ such that
$B_{n} \frac{a_{n}}{a_{n+1}}-B_{n+1} \leq 0$ for all $n \geq N$. Rearranging this inequality gives us
$$
B_{n} a_{n} \leq B_{n+1} a_{n+1}
$$
for all $n \geq N$ which then implies that
$$
B_{N} a_{N} \leq B_{n} a_{n}
$$
for all $n \geq N$. Letting $B_{N} a_{N}$ be a constant $C$, we have
$$
a_{n} \geq \frac{C}{B_{n}}
$$
for $n \geq N$ and, because $\sum_{n=1}^{\infty} \frac{1}{B_{n}}$ diverges, the series $\
sum_{n=1}^{\infty} a_{n}$ diverges by the Comparison Test.}
\begin{enumerate}
\setcounter{enumi}{1}
\item Let $\left(u_{n}\right)_{n \geq 0}$ be a sequence of positive real numbers satisfying
\end{enumerate}
$$
\lim _{n \rightarrow+\infty} n \ln \left(\frac{u_{n}}{u_{n+1}}\right)=\ell, \quad-\infty \leq \ell \leq+\
infty
$$
Show that the series $\sum_{n \geq 0} u_{n}$ converges if $\ell>1$ and diverges if $\ell<1$.
Proof : we have $: \forall n \geqslant 1:\left(1+\frac{1}{n}\right)^{n} \leqslant e \leqslant\left(1+\
frac{1}{n}\right)^{n+1}$
where $\forall n \geqslant 1: \frac{1}{n+1}<\ln \left(1+\frac{1}{n}\right)<\frac{1}{n}$
\begin{enumerate}
\item if $l>1, \exists \varepsilon>0$, such that $\rho=l-\varepsilon>1$ and for all $n \in \mathbb{N}
$, on $a \forall n \geqslant N$ $n \log \frac{u_{n}}{u_{n+1}}>\rho>\rho n \ln \left(1+\frac{1}{n}\right)
$ then $\ln \frac{u_{n}}{u_{n+1}}>\ln \frac{(n+1)^{\rho}}{n^{\rho}}$, $\forall n \geqslant N \
Longleftrightarrow \frac{u_{n}}{u_{n+1}}>\frac{(n+1)^{\rho}}{n^{\rho}} \Longleftrightarrow \
frac{u_{n+1}}{u_{n}} \leqslant \frac{n^{\rho}}{(n+1)^{\rho}}=\frac{\frac{1}{(n+1)^{\rho}}}{\frac{1}{n^{\
rho}}}$ and since $\sum_{n=1}^{+\infty} \frac{1}{n^{\rho}}(\rho>1)$ converge then the serie $\
sum_{n=1}^{+\infty} u_{n}$ converge.
\item if $l<1$ then there exist $\varepsilon>0$, such that $\rho=l+\varepsilon<1$ and for all $N \in \
mathbb{N}$, we have $\forall n \geqslant N: n \ln \frac{u_{n}}{u_{n+1}}<l+\varepsilon<1, \forall n \
geqslant N: n \ln \frac{u_{n}}{u_{n+1}}<n \ln \left(1+\frac{1}{n-1}\right)$
\end{enumerate}
$\Longleftrightarrow \forall n \geqslant N \frac{u_{n}}{u_{n+1}}<\frac{n}{n-1} \Longleftrightarrow \
forall n \geqslant N, \frac{u_{n+1}}{u_{n}}>\frac{n-1}{n}=\frac{\frac{1}{n}}{\frac{1}{n-1}}$
and since the serie $\sum_{n=2}^{+\infty} \frac{1}{n-1}$ diverge then $\sum_{n=0}^{+\infty} u_{n}$
diverge.
$$
\begin{enumerate}
\setcounter{enumi}{2}
3. \item (Raabe's Test) Let $\left(u_{n}\right)_{n \geq 0}$ be a sequence of positive real numbers
satisfying
\end{enumerate}
$$
\lim _{n \rightarrow+\infty} n\left(\frac{u_{n}}{u_{n+1}}-1\right)=\ell
$$
Show that the series $\sum_{n \geq 0} u_{n}$ converges if $\ell>1$ and diverges if $\ell<1$.
$$
Proof. Items (i) and (ii) are consequences of Kummer's Test. To demonstrate this, let $1/d_{n}=n$
and compute
$$
\begin{aligned}
L & =\lim _{n \rightarrow \infty}\left(n \frac{u_{n}}{u_{n+1}}-(n+1)\right) \\
& =\lim _{n \rightarrow \infty}\left(n\left(\frac{u_{n}}{u_{n+1}}-1\right)-1\right) \\
& =\rho-1
\end{aligned}
$$
If $\rho>1$, then $L>0$ and the series $\sum_{n=1}^{\infty} a_{n}$ converges by Kummer's Test.
Similarly, if $\rho<1$, then $L<0$ and note that $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, thus the
series $\sum_{n=1}^{\infty} u_{n}$ diverges by Kummer's Test.
$$
\begin{enumerate}
\setcounter{enumi}{3}
4. \item (Gauss's Test) Let $\left(a_{n}\right)_{n \geq 0}$ be a sequence of positive real numbers
satisfying
\end{enumerate}
$$
\frac{a_n}{a_{n+1}}=\lambda+\frac{\beta}{n}+\frac{\gamma_{n}}{n^{1+\varepsilon}}, \quad \text
{ for } n \text { sufficiently large. }
$$
where $\lambda, \beta, \varepsilon>0$ are real constants and $\left|\theta_{n}\right| \leq M, M>0$.
Show that
(a) The series $\sum_{n \geq 0} u_{n}$ converges if $\lambda>1$ or $(\lambda=1$ and $\beta>1)$.
(b) The series $\sum_{n \geq 0} u_{n}$ diverges if $\lambda<1$ or $(\lambda=1$ and $\beta \leq
1)$.\\
Proof
a) $\lambda \neq 1 \Rightarrow \frac{a_n}{a_{n +1}}=\lambda+\frac{\beta}{n}+\frac{\gamma_n}
{n^{1+\varepsilon}} ; \lim _{n \rightarrow +\infty} \frac{a_n}{a_{n+1}}=\lambda$
Using Alembert Ratio test:
- $\sum a_n$ converge if $\lambda>1$
- $\sum a_n$ diverge if $\lambda<1$
b) $\lambda=1 \Rightarrow \frac{a_n}{a_{n+1}}=1+\frac{\beta}{n}+\frac{\gamma_n}{n^{1+\
varepsilon}}$
$$
\begin{gathered}
n\left(\frac{a_n}{a_{n+1}}-1\right)=\beta +\frac{\gamma_n}{n^{{\varepsilon}}} ;\lim _{n \rightarrow
+\infty} n\left(\frac{a_n}{a_{n+1}}-1\right)=\beta\\
\end{gathered}
$$
using Raabe's test:
- $\sum a_n$ converge if $\beta>1$
- $\sum a_n$ diverge if $\beta<1$
- if $\beta=1$;
$$
\begin{gathered}
n\left(\frac{a_n}{a_{n+1}}-1\right)-1=\frac{\gamma_n}{n^{\varepsilon}} \\
\lim _{n \rightarrow+\infty}\left[n\left(\frac{a_n}{a_{n+1}}-1\right )-1\right ] \log n=\lim _{n \
rightarrow+\infty} \frac{\gamma_n}{n^{\varepsilon}} \log n=0<1
\end{gathered}
$$
using bertland test $\sum a_n$ deverge
\begin{enumerate}
\setcounter{enumi}{4}
\item (Duhamel's Test) Let $\left(u_{n}\right)_{n \geq 0}$ be a sequence of positive real numbers
satisfying
\end{enumerate}
$$
\frac{u_{n+1}}{u_{n}}=1-\frac{\beta}{n}+o\left(\frac{1}{n}\right), \quad \text { as } n \rightarrow+\
infty
$$
where $\beta$ is a constant. Show that the series $\sum_{n \geq 0} u_{n}$ converges if $\beta>1$
and diverges if $\beta<1$.
$$
Let $v_{n}=\frac{1}{n^{\alpha}}$.
we have :
$$
\frac{u_{n+1}}{u_{n}}=1-\frac{\beta}{n}+o\left(\frac{1}{n}\right)
$$
Then :
LET $\left(u_{n}\right)$ a sequence of positive terms, satisfy $\frac{u_{n+1}}{u_{n}}=1-\frac{\beta}{n}
+o\left(\frac{1}{n}\right), \beta C^{t e}$
If $\beta>1$, then $\sum_{n=0}^{+\infty} u_{n}$ converge.
IF $\beta<1$, then $\sum_{n=0}^{+\infty} u_{n}$ diverge.
Proof : let $\alpha$ a real number, and $v_{n}=\frac{1}{n^{\alpha}}$, we get $\frac{v_{n+1}}{v_{n}}-\
frac{u_{n+1}}{u_{n}}=$ $\frac{\beta-\alpha}{n}+o\left(\frac{1}{n}\right)$. Si $\beta \neq \alpha$, then
the difference $\frac{v_{n+1}}{v_{n}}-\frac{u_{n+1}}{u_{n}}$ has the same sign of $\beta-\alpha$
if $\beta>1$, we can choose $\alpha$ such that $\beta>\alpha>1$ the serie $\sum_{n=0}^{+\infty}
v_{n}$ is convergent and $\frac{u_{n+1}}{u_{n}} \leqslant \frac{v_{n+1}}{v_{n}}$ show that the serie
$\sum_{n=0}^{+\infty} u_{n}$ is convergent.
if $\beta<1$, we can choose $\alpha$ such that $\beta<\alpha<1$ the serie $\sum_{n=0}^{+\infty}
v_{n}$ is divergent and $\frac{u_{n+1}}{u_{n}} \geqslant \frac{v_{n+1}}{v_{n}}$ show that the serie
$\sum_{n=0}^{+\infty} u_{n}$ diverge.
$$
\begin{enumerate}
\setcounter{enumi}{5}
\item (Schlömilch Condensation Test) Let $\left(u_{n}\right)_{n \geq 1}$ be a decreasing sequence
of positive real numbers, and let $\left(n_{k}\right)_{k \geq 1}$ be an increasing sequence of positive
integers such that
\end{enumerate}
$$
\exists C>0, \forall k \geq 1: \frac{n_{k+1}-n_{k}}{n_{k}-n_{k-1}} \leq C
$$
Show that the series $\sum_{n \geq 1} u_{n}$ converges if and only if the series $\sum_{k \geq 1}\
left(n_{k+1}-n_{k}\right) u_{n_{k}}$ converges.\\
PROOF:
Since $(a_i)$ is non-increasing, one has
\[
(n_{k+1}-n_k)a_{n_k} \geqslant \sum_{i=n_k+1}^{n_{k+1}}a_i.
\]
Moreover, using the assumption on $(n_k)$,
\[
a_{n_k}(n_{k+1}-n_k) \leqslant Ca_{n_k}(n_k-n_{k-1}) \leqslant C\sum_{i=n_{k-1}+1}^{n_k} a_i.
\]
Denoting $S_N:=\sum_{i=n_0+1}^Na_i$, one has, in view of the previous inequalities, that
\[
S_{n_{K+1}} \leqslant \sum_{k=1}^Ka_{n_k}(n_{k+1}-n_k) \leqslant CS_{n_K}.
\]
\section*{Problems}
\begin{enumerate}
\item (Riemann Theorem) Let $\sum_{n \geq 0} a_{n}$ be a conditionally convergent series.
\end{enumerate}
\begin{itemize}
\item Show that for any $B \in \mathbb{R}$, there exists a permutation $\sigma$ of $\mathbb{N}$
such that
\end{itemize}
$$
\sum_{n=0}^{+\infty} a_{\sigma(n)}=B
$$
\begin{itemize}
\item Show also that there exists a permutation $\sigma$ of $\mathbb{N}$ such that
\end{itemize}
$$
\sum_{n=0}^{+\infty} a_{\sigma(n)}=+\infty
$$
Riemann Rearrangement Theorem: Given a conditionally convergent real series
$$
\sum_{n=1}^{\infty} a_n
$$
and a value $M \leqslant \mathbb{R}$, there exists a rearrangement of the veries such that $\sum
a_{\sigma(n)}=M$. Proof. Given $\sum a_n$ is conditionally convergent, $\sum\left|a_n\right|=\
infty$. Define subsequences ${ }^1\left(a_{n_j}\right)_{n, \in A}$ and $\left(a_{n_k}\right)_{m_k \in
H}$ of $a_n$ by $i \in A \Leftrightarrow a_i<0$ and $i \in B \Leftrightarrow a_i \geq 0$. Claim: $\
sum_{j=1}^{\infty} a_{n_1}=-\infty$ and $\sum_{k=1}^{\infty} a_{n_1}=\infty$. Suppose both series
converge. Then by scries addition $\sum\left|a_n\right|=\sum_{k=1}^{\infty} a_{n d}-\sum_{j=1}^{\
infty} a_{n j}$ converges. A contradiction. Suppowe one scries converges and the other scries
diverges, Then $\sum_{k=1}^{\infty} a_{n_k}+\sum_{j=1}^{\infty} a_{n_j}=\sum a_n$ diverges,
Another contradiction. Now for the construction of permutation $\sigma$ of $\mathbb{N}$. Let
$j_1$ be the smallest $\mathbb{N}$ such that
$$
\sum_{j=1}^j a_{n j}<M .
$$
Define $\sigma(j)=n_j \in A, \forall j \in\left[1 . . j_1\right] .^2$ Let $k_1$ be the smallest $\
mathbb{N}$ such that
$$
\sum_{j=1}^{j_1} a_{n_j}+\sum_{k=1}^{k_z} a_{n_k}>M
$$
Detine $\left.\sigma\left(j_1+k\right)=n_k \in B . \forall k \in \mid 1 . . k_1\right\}$. Step 2: Let $j_2$
be the smallest $\mathrm{N}$ such that
$$
\sum_{j=1}^h a_{n,}+\sum_{k=1}^{k_2} a_{n_k}<M
$$
Define $\sigma\left(j+k_1\right)=n_j \in A, \forall j \in(j, \cdot j)$. Let $k_2$ be the smallest $\
mathbb{N}$ such that
$$
\sum_{j=1}^{j_2} a_{n_j}+\sum_{k=1}^{k_3} a_{n_k}>M .
$$
'Is there a less cumbersocue may to define these sulesequesces? ${ }^2$ Here the nolation [a.b]
refers to all the integres from a through b. Also $(a, b)$ is the set of all iscegers between a asd b. 1
Define $\sigma\left(j_2+k\right)=n_k \in B \cdot \forall k \in\left(k_1 \cdot k_2\right)$. Continue
defining $\sigma$ as above and it will be a permutation of $\mathrm{N}$ such that the series
rearrangement $\sum a_{g(n)}$ will continue to oscillate around $M$. First by summing, in order, the
negative terms from the sequence $\left(a_n\right)$ until the last negative term drops it below $M$.
Then by adding to the sum, in order, from the non-negative terms of sequence $\left(a_n\right)$
until the last term pushes is over $M$.
Let $c>0$. By the divergence test $\left|a_n\right| \rightarrow 0$. Thas $\exists N \in N$ such that
$\forall n \geq N\left|a_n\right|<\varepsilon$. Now $\exists i \in \mathbb{N}$ such that $j_i+k_i>N$.
Then since
$$
\sum_{j=1}^{N_1} a_{n_j}+\sum_{k=1}^{k_j} a_{n k}>M \geq \sum_{j=1}^{j_k} a_{n_j}+\
sum_{k=1}^{N_j-1} a_{n k}
$$
we bave $\forall p \geq j+k_i\left|M-\sum_{n=1}^p a_{\rho(n)}\right|<\varepsilon$. Therefore $\sum
a_{\rho(n)}=M$.
\begin{enumerate}
\setcounter{enumi}{1}
\item (Kempner series) Let $\left(u_{n}\right)_{n \geq 1}$ be the sequence of all positive integers
whose decimal representation does not contain the digit 9 . Show that the series $\sum_{n \geq 1} \
frac{1}{u_{n}}$ is convergent.
$$
we have
$\sum_{n=0}^{+\infty} 1/u_{n}$=$$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{8}+\frac{1}{10}+\frac{1}
{11}+\cdots+\frac{1}{18}+\frac{1}{20}+\frac{1}{21}+\cdots+\frac{1}{87}+\frac{1}{88}+\frac{1}{100}+\
frac{1}{101}+\cdots=$$=$$
\left(1+\frac{1}{2}+\cdots+\frac{1}{8}\right)+\left(\frac{1}{10}+\cdots+\frac{1}{88}\right)+\left(\
frac{1}{100}+\cdots+\frac{1}{888}\right)+\cdots$$=$$
\begin{aligned}
& (\overbrace{1+\cdots+\frac{1}{8}}^{8 \text { terms }})+(\overbrace{\frac{1}{10}+\cdots+\frac{1}
{88}}^{8 \times 9 \text { terms }})+(\overbrace{\frac{1}{100}+\cdots+\frac{1}{888}}^{8 \times 9^{2} \
text { terms }})+\cdots+(\overbrace{\frac{1}{10^{n}}+\cdots+\frac{1}{\underbrace{88 \ldots 8}_{n-\
text { times }}}}^{8 \times 9^{n-1} \text { terms }})+\ldots \\
& <(\overbrace{1+\cdots+1}^{8 \text { terms }})+(\overbrace{\frac{1}{10}+\cdots+\frac{1}{10}}^{8 \
times 9 \text { terms }})+(\overbrace{\frac{1}{100}+\cdots+\frac{1}{100}}^{8 \times 9^{2}})+\cdots+(\
overbrace{\frac{1}{10^{n}}+\cdots+\frac{1}{10^{n}}}^{8 \times 9^{n-1} \text { terms }})+\ldots \\
& =8+\frac{8 \times 9}{10}+\frac{8 \times 9^{2}}{10^{2}}+\frac{8 \times 9^{3}}{10^{3}}+\ldots+\frac{8
\times 9^{n}}{10^{n}}+\ldots=8 \times \frac{1}{1-(9 / 10)}=80
\end{aligned}
$$
then the serie $\sum_{n=0}^{+\infty} 1/u_{n}$ is convergent.
$$
3. \item Let $\left(a_{n}\right)_{n \geq 0}$ be a decreasing sequence of positive numbers with limit
0 , and let
\end{enumerate}
$$
b_{n}=a_{n}-2 a_{n+1}+a_{n+2} \geq 0, \quad \forall n \geq 0
$$
Show that
$$
\sum_{n=0}^{+\infty} n b_{n}=a_{1}
$$
for all $n>2$ An appears in three terms:
$\mathrm{Bn}, \mathrm{B}(\mathrm{n}-1), \mathrm{B}(\mathrm{n}-2)$
when we evaluate them sum:
$n B n+(n-1) B(n-1)+(n-2) B(n-2)$
$=n A n-2 n A(n+1)+n A(n+2)+(n-1) A(n-1)-2(n-1) A(n)+(n-1) A(n+1)+(n-2) A(n-2)-2(n-2) A(n-1)+(n-2)$
$A(n)$
the An appears in nAn, -2(n-1)An , (n-2) An
when we sum them we get 0 hence for all $n>2$ the Anth term disappear after the summation
for $\mathrm{n}=2$ it goes with 0 too because it appears in two terms
when we sum $O B(0)+B(1)+2 B(2)$ we get : $A(1)-2 A(2)+A(3)+2 A(2)-4 A(3)+A(4)$ where the term
$A(2)$ disappears too
for $n=0, A(0)$ is already multiplied by 0 so it disappears too
hence all the terms disappear except $\mathrm{A}(1)$ because it only appears once
we get the result
\begin{enumerate}
\setcounter{enumi}{4}
4. \item Evaluate the double series
\end{enumerate}
$$
\sum_{j=0}^{+\infty} \sum_{k=0}^{+\infty} 2^{-3 k-j-(k+j)^{2}}
$$
\begin{enumerate}
\setcounter{enumi}{4}
\item Let $\left(u_{n}\right)_{n \geq 1}$ be a sequence of positive integers. For an integer $m$ let
$V_{m}$ denote the number of terms of $\left(u_{n}\right)_{n \geq 1}$ that are less than or equal to
$m$. Show that if $\sum_{n \geq 1} 1 / u_{n}$ is convergent, then
\end{enumerate}
$$
\lim _{n \rightarrow \infty} \frac{V_{n}}{n}=0
$$
\textbf{Proof:}
1. First, note that since \( (u_n)_{n\geq 1} \) is a sequence of positive integers, \( V_m \) is a non-
decreasing sequence.
2. Consider the partial sums \( S_k = \sum_{n=1}^{k} \frac{1}{u_n} \). The given condition implies that
\( S_k \) is a convergent sequence.
3. Let \( N_m \) be the number of terms in \( (u_n)_{n\geq 1} \) that are less than or equal to \( m \).
Then, \( V_m = N_m \) since \( V_m \) is defined as the number of terms up to \( m \).
4. Now, \( \frac{V_n}{n} = \frac{N_n}{n} \). Observe that \( N_n \leq m \) for all \( n \) since every
term in \( (u_n)_{n\geq 1} \) is less than or equal to \( m \) for \( n \geq N_m \).
5. Therefore, \( \frac{V_n}{n} \leq \frac{m}{n} \). As \( n \) approaches infinity, the right-hand side
goes to zero, implying \( \lim_{n\to\infty} \frac{V_n}{n} = 0 \).
This completes the proof, showing that if \( \sum_{n=1}^{\infty} \frac{1}{u_n} \) is convergent, then \(
\lim_{n\to\infty} \frac{V_n}{n} = 0 \).
\begin{enumerate}
\setcounter{enumi}{5}
\item Let $f_{0}(x)=e^{x}$ and
\end{enumerate}
$$
f_{n+1}(x)=x f_{n}^{\prime}(x), \quad n=0,1,2, \ldots
$$
Show that
$$
\sum_{n=0}^{+\infty} \frac{f_{n}(1)}{n !}=e^{e}
$$
$$
Solution. We start by exploring the cases, we have
$$
\begin{aligned}
& f_{1}(x)=(x) \\
& f_{2}(x)=\left(x^{2}+x\right) \\
& f_{3}(x)=\left(x^{3}+3 x^{2}+x\right) \\
& f_{4}(x)=\left(x^{4}+6 x^{3}+7 x^{2}+x\right) \\
& f_{5}(x)=\left(x^{5}+10 x^{4}+25 x^{3}+15 x^{2}+x\right)
\end{aligned}
$$
We conjecture and proof by induction that $\forall n \geq 1, \exists P_{n} \in \mathbb{R}[X], f_{n}
(x)=$ $P_{n}(x) e^{x}$. There are enough base cases in the exploration part, now we prove the step.
We take as a hypothesis that $f_{n}(x)=P_{n}(x) e^{x}$ and we prove that $f_{n+1}(x)=P_{n+1}(x)
e^{x}$
$$
\begin{aligned}
f_{n+1}(x) & =x f^{\prime}(x)=x\left[P_{n}(x) e^{x}\right]^{\prime} \\
& =x\left[P_{n}^{\prime}(x) e^{x}+P_{n}(x) e^{x}\right] \\
& =\left[x P_{n}^{\prime}(x)+x P_{n}(x)\right] e^{x} \\
& =P_{n+1}(x) e^{x}
\end{aligned}
$$
We can clearly see that $x P_{n}^{\prime}(x)+x P_{n}(x)$ is a polynomial. We have $P_{1}(x)=x$ and
$P_{n+1}(x)=x P_{n}^{\prime}(x)+x P_{n}(x)$. Now we find an explicit formula for $P_{n}$ that is using
its coefficents (we can clearly see that $P_{n}$ degree is $n$ ), we have
$$
P_{n}(x)=\sum_{k=1}^{n} a_{n, k} x^{k} \quad \Longrightarrow \quad P_{n}^{\prime}(x)=\
sum_{k=1}^{n-1}(k+1) a_{n, k+1} x^{k}
$$
Given that $\operatorname{deg} P_{n}=n$ then $\forall k>n, a_{n, k}=0$ Thus getting
$$
\begin{aligned}
P_{n+1}(x) & =x\left[P_{n}^{\prime}(x)+P_{n}(x)\right] \\
& =x\left[\sum_{k=1}^{n-1}(k+1) a_{n, k+1} x^{k}+\sum_{k=1}^{n} a_{n, k} x^{k}\right]=x\left[\
sum_{k=1}^{n}(k+1) a_{n, k+1} x^{k}+\sum_{k=1}^{n} a_{n, k} x^{k}\right] \\
& =\sum_{k=1}^{n}\left[(k+1) a_{n, k+1}+a_{n, k}\right] x^{k}=\sum_{k=1}^{n} a_{n+1, k} x^{k}
\end{aligned}
$$
Thus getting $a_{n+1, k}=(k+1) a_{n, k+1}+a_{n, k}$ which is the formula for Stirling numbers. Also it
results that $P_{n}(x)=T_{n}(x)$ where $T_{n}(x)$ is the $\mathrm{n}^{\text {th }}$ Touchard
polynomial. We have that $T_{n}(1)=\sum_{k=1}^{n} a_{n, k}=B_{n}$ where $B_{n}$ are the Bell
numbers. We have an important recursion formula which is $B_{n+1}=$
$\sum_{k=1}^{n}\left(\begin{array}{c}n \\ k\end{array}\right) B_{k}$. We define $B(x)=\sum_{n \geq
0} \frac{B_{n}}{n !} x^{n}$. Now we try to find an explicit formula for $B(x)$. The easiest way to go is
to solve it using a differential equation so we have
$$
\begin{aligned}
B^{\prime}(x) & =\sum_{n \geq 1} \frac{B_{n}}{(n-1) !} x^{n-1}=\sum_{n \geq 0} \frac{B_{n+1}}{n !}
x^{n} \\
& =\sum_{n \geq 0} \frac{x^{n}}{n !} \sum_{k=0}^{n}\left(\begin{array}{c}
n \\
\end{array}\right) B_{k}=\sum_{n \geq 0} \sum_{k=0}^{n} \frac{x^{n-k}}{(n-k) !} \frac{x^{k}}{k !}
B_{k} \\
& =\sum_{k \geq 0} \frac{x^{k}}{k !} B_{k} \sum_{n=k}^{\infty} \frac{x^{n-k}}{(n-k) !}=\sum_{k \geq
0} \frac{x^{k}}{k !} B_{k} \sum_{n=0}^{\infty} \frac{x^{n}}{n !} \\
& =e^{x} \sum_{k \geq 0} \frac{x^{k}}{k !} B_{k}=e^{x} \cdot B(x)
\end{aligned}
$$
Therefore we get that $B(x)$ is the solution of the differential equation $y^{\prime}=$ $y e^{x}$ with
$y(0)=1$. We solve the differential system
$$
\begin{aligned}
y^{\prime}=y e^{x} & \Longrightarrow \frac{d y}{d x}=y e^{x} \\
& \Longrightarrow \frac{d y}{y}=e^{x} d x \\
& \Longrightarrow \int \frac{d y}{y}=\int e^{x} d x \\
& \Longrightarrow \ln (y)=e^{x}+C \\
& \Longrightarrow y=e^{e^{x}+C}
\end{aligned}
$$
Since $y(0)=1$ then $e^{1+C}=1$ so $1+C=0$ giving $C=-1$. We get the result $B(x)=e^{e^{x}-1}$. All
thats left now is to evaluate the wanted sum
$$
\sum_{n \geq 0} \frac{f_{n}(1)}{n !}=\sum_{n \geq 0} \frac{e \cdot B_{n}}{n !}=e \sum_{n \geq 0} \
frac{B_{n} \cdot 1^{n}}{n !}=e B(1)=e \cdot e^{e^{1}-1}=e^{e}
$$
\begin{enumerate}
\setcounter{enumi}{6}
\item For any integer $n \geq 1$, let $\langle n\rangle$ denote the closest integer to $\sqrt{n}$.
Evaluate
\end{enumerate}
$$
\sum_{n=1}^{+\infty} \frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}{2^{n}}
$$
$$
\section*{Part (a)}
My approach: It is clear that there are exactly $2n$ integers between $(n+1)^{2}$ and $n^{2}$ for all
$n \in \mathbb{N}$. Now, by the pattern that we observe, we can conclude that out of these $2n$
integers, the first $n$ integers, let us call them $k$, have $\langle k\rangle=n$, and the other $n$
integers, let us call them $l$, have $\langle l\rangle=n+1$.
Therefore, for any $k \in \mathbb{N}$, the positive integers $n$ that have $\langle n\rangle=k$ are
\[
\begin{aligned}
& k^{2}-(k-1), k^{2}-(k-2), \ldots, k^{2} \\
& k^{2}+1, k^{2}+2, \ldots, k^{2}+k.
\end{aligned}
\]
Therefore, we are done with part (a) of the question.
\section*{Part (b)}
Now, for any $k \in \mathbb{N}$, we have
\[
\begin{aligned}
& \sum_{n=k^{2}-(k-1)}^{k^{2}+k} \frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}{2^{n}} \\
& =\sum_{n=k^{2}-(k-1)}^{k^{2}+k} \frac{2^{k}+2^{-k}}{2^{n}} \\
& =\left(2^{k}+2^{-k}\right) \sum_{n=k^{2}-(k-1)}^{k^{2}+k} \frac{1}{2^{n}} \\
& =\left(2^{k}+2^{-k}\right) \frac{2^{-\left(k^{2}-(k-1)\right)}\left(\left(2^{-1}\right)^{2 k}-1\right)}
{2^{-1}-1} \\
& =\left(2^{k}+2^{-k}\right) \frac{2^{k-1-k^{2}}\left(2^{-2 k}-1\right)}{2^{-1}-1} \\
& =\left(2^{k}+2^{-k}\right) \frac{2^{-k^{2}-k-1}-2^{k-1-k^{2}}}{2^{-1}-1} \\
& =\left(2^{k}+2^{-k}\right)\left(2^{k-k^{2}}-2^{-k^{2}-k}\right) \\
& =2^{2 k-k^{2}}-2^{-k^{2}-2 k}=2^{1-(k-1)^{2}}-2^{1-(k+1)^{2}} \\
& =\left(2^{1-(k-1)^{2}}+2^{1-k^{2}}\right) - \left(2^{1-k^{2}}+2^{1-(k+1)^{2}}\right).
\end{aligned}
\]
Define the sequence $\{a_{n}\}_{n\geq 1}$ such that
\[
a_{n}=2^{1-(n-1)^{2}}+2^{1-n^{2}}, \quad \forall n \in \mathbb{N}.
\]
Therefore, for all $k \in \mathbb{N}$, we have
\[
\sum_{n=k^{2}-(k-1)}^{k^{2}+k} \frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}{2^{n}}=a_{k}-
a_{k+1}.
\]
This implies that
\[
\begin{aligned}
\sum_{n=1}^{\infty} \frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}{2^{n}} &= \sum_{i=1}^{\infty}
(a_{i}-a_{i+1}) \\
&= \lim_{l \rightarrow \infty} (a_{1}-a_{l+1}) \\
&= 3.
\end{aligned}
\]
Hence,
\[
\sum_{n=1}^{\infty} \frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}{2^{n}}=3.
\]
\end{document}
