The Maclaurin series for a function $f$ is given by $\sum_{n=1}^{\infty} \frac{(n+1)x^n}{n^2 6^n}$ and converges to $f(x)$ for all $x$ in the interval of convergence. It can be shown that the Maclaurin series for $f$ has a radius of convergence of 6.
(a) Determine whether the Maclaurin series for $f$ converges or diverges at $x = 6$. Give a reason for your answer.
(b) It can be shown that $f(-3) = \sum_{n=1}^{\infty} \frac{(n+1)(-3)^n}{n^2 6^n} = \sum_{n=1}^{\infty} \frac{n+1}{n^2}\left(-\frac{1}{2}\right)^n$ and that the first three terms of this series sum to $S_3 = -\frac{125}{144}$. Show that $\left|f(-3) - S_3\right| < \frac{1}{50}$.
(c) Find the general term of the Maclaurin series for $f'$, the derivative of $f$. Find the radius of convergence of the Maclaurin series for $f'$.
(d) Let $g(x) = \sum_{n=1}^{\infty} \frac{(n+1)x^{2n}}{n^2 3^n}$. Use the ratio test to determine the radius of convergence of the Maclaurin series for $g$.

Show that a power series $\sum _ { n \geq 0 } a _ { n } z ^ { n }$ where $a _ { n } \rightarrow 0$ as $n \rightarrow \infty$ cannot have a pole on the unit circle. Is the statement true with the hypothesis that $\left( a _ { n } \right)$ is a bounded sequence?

Let $\left( a _ { n } \right) _ { n \in \mathbb { N } }$ be a complex sequence. We assume that the series $\sum a _ { n }$ converges. For $n \in \mathbb { N }$, we denote $r _ { n } = \sum _ { k = n + 1 } ^ { + \infty } a _ { k }$ and we define the functions $s _ { n }$ and $s$ from $[ 0,1 ]$ to $\mathbb { C }$ by $s _ { n } ( x ) = \sum _ { k = 0 } ^ { n } a _ { k } x ^ { k }$ and $s ( x ) = \sum _ { k = 0 } ^ { + \infty } a _ { k } x ^ { k }$.
Justify the existence of $s$.

Let $\left( a _ { n } \right) _ { n \in \mathbb { N } }$ be a complex sequence. We assume that the series $\sum a _ { n }$ converges. For $n \in \mathbb { N }$, we denote $r _ { n } = \sum _ { k = n + 1 } ^ { + \infty } a _ { k }$ and we define the functions $s _ { n }$ and $s$ from $[ 0,1 ]$ to $\mathbb { C }$ by $s _ { n } ( x ) = \sum _ { k = 0 } ^ { n } a _ { k } x ^ { k }$ and $s ( x ) = \sum _ { k = 0 } ^ { + \infty } a _ { k } x ^ { k }$.
Let $x \in [ 0,1 ]$ and $n \in \mathbb { N } ^ { * }$. Show $$s ( x ) - s _ { n } ( x ) = r _ { n } x ^ { n + 1 } - \sum _ { k = n + 1 } ^ { + \infty } r _ { k } \left( x ^ { k } - x ^ { k + 1 } \right)$$

Let $\left( a _ { n } \right) _ { n \in \mathbb { N } }$ be a complex sequence. We assume that the series $\sum a _ { n }$ converges. For $n \in \mathbb { N }$, we denote $r _ { n } = \sum _ { k = n + 1 } ^ { + \infty } a _ { k }$ and we define the functions $s _ { n }$ and $s$ from $[ 0,1 ]$ to $\mathbb { C }$ by $s _ { n } ( x ) = \sum _ { k = 0 } ^ { n } a _ { k } x ^ { k }$ and $s ( x ) = \sum _ { k = 0 } ^ { + \infty } a _ { k } x ^ { k }$.
Show that $s$ is continuous on $[ 0,1 ]$.
For continuity at 1, fix $\varepsilon > 0$ and show that if the natural integer $N$ satisfies $\left| r _ { n } \right| \leqslant \varepsilon$ for all $n \geqslant N$, then $\left| s ( x ) - s _ { N } ( x ) \right| \leqslant 2 \varepsilon$ for all $x \in [ 0,1 ]$. Then bound the modulus of $s ( x ) - s ( 1 ) = \left( s ( x ) - s _ { N } ( x ) \right) + \left( s _ { N } ( x ) - s _ { N } ( 1 ) \right) + \left( s _ { N } ( 1 ) - s ( 1 ) \right)$.

Let $\left( a _ { n } \right) _ { n \in \mathbb { N } }$ be a complex sequence. We assume that the series $\sum a _ { n }$ converges. For $n \in \mathbb { N }$, we denote $r _ { n } = \sum _ { k = n + 1 } ^ { + \infty } a _ { k }$ and we define the functions $s _ { n }$ and $s$ from $[ 0,1 ]$ to $\mathbb { C }$ by $s _ { n } ( x ) = \sum _ { k = 0 } ^ { n } a _ { k } x ^ { k }$ and $s ( x ) = \sum _ { k = 0 } ^ { + \infty } a _ { k } x ^ { k }$.
Application: recover the result from question II.B.3.

Let $\theta \in \mathbb { R }$. Determine the power series expansion of the function $$x \mapsto \frac { 1 - x ^ { 2 } } { x ^ { 2 } - 2 x \cos \theta + 1 }$$ on an interval to be specified.

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a $2 \pi$-periodic function of class $\mathcal { C } ^ { 1 }$. We consider the Fourier series of $f$ in cosines and sines, denoted $$c _ { 0 } + \sum _ { n \geq 1 } \left( a _ { n } \cos ( n t ) + b _ { n } \sin ( n t ) \right)$$
Show that, for all $x \in ] - 1,1 [$ and all $t \in \mathbb { R }$, $$c _ { 0 } + \sum _ { n = 1 } ^ { + \infty } \left( a _ { n } \cos ( n t ) + b _ { n } \sin ( n t ) \right) x ^ { n } = \frac { 1 } { 2 \pi } \int _ { 0 } ^ { 2 \pi } \frac { \left( 1 - x ^ { 2 } \right) f ( u ) } { x ^ { 2 } - 2 x \cos ( t - u ) + 1 } \mathrm { ~d} u$$

We use the notation $R$ introduced in part I and $V_n(z) = U_{n+1}(z,-1)$. Let $z \in \mathbb{C}$ such that $z^2 \neq 1$, with $r$, $s$, $t$, $h$ as defined in II.C.1. What can be said about the radius of convergence of the power series $Z \mapsto \sum_{n=0}^{+\infty} V_n(z) Z^n$? We denote $g_z$ its sum.

We use the notation $V_n(z) = U_{n+1}(z,-1)$ and $g_z$ the sum of the power series $\sum_{n=0}^{+\infty} V_n(z) Z^n$. When it makes sense, calculate $\left(1 - 2zZ + Z^2\right) g_z(Z)$.

Let $z \in \mathbb{C}$. We denote $\Omega_z$ the set of points in the plane with complex affixe $Z$ such that $|Z(Z-2z)| < 1$, and $V_n(z) = U_{n+1}(z,-1)$. Show that there exists a non-empty open disk $\Delta$ with center $O$ included in $\Omega_z$ such that $$\forall Z \in \Delta, \quad \frac{1}{1 - 2zZ + Z^2} = \sum_{n=0}^{+\infty} V_n(z) Z^n = \sum_{p=0}^{+\infty} \left(Z^p(2z - Z)^p\right)$$

We assume $\alpha = 1$ and use the notation $V_n(z) = U_{n+1}(z,-1)$. Let $t \in ]0,\pi[$. Show that the function $$H_t : x \mapsto \frac{1}{1 - 2x\cos(t) + x^2}$$ is expandable as a power series on $]-1,1[$.

We consider a function $f$ of class $\mathcal{C}^2$ on $[0,1]$ taking values in $[0,1]$ such that $f'$ and $f''$ take non-negative values. We assume $f(1)=1$, $f'(0)<1$ and $f''(1)>0$. We set $m=f'(1)$. We consider the recurrent sequence $(u_n)_{n\in\mathbb{N}}$ defined by $u_0=0$ and, for all $n\in\mathbb{N}$, $u_{n+1}=f(u_n)$.
We now assume $m<1$ and we set again, for $n\in\mathbb{N}$, $\varepsilon_n=1-u_n$.
Show that the series with general term $\varepsilon_n$ is absolutely convergent and deduce the convergence of the series with general term $\ln\left(\frac{m^{-(n+1)}\varepsilon_{n+1}}{m^{-n}\varepsilon_n}\right)$.

A function $f$, defined on $D(0,R) \subset \mathbb{R}^2$ and with complex values, expands as a power series on $D(0,R)$ if there exists a complex sequence $(a_n)$ such that $$\forall (x,y) \in D(0,R), \quad f(x,y) = \sum_{n=0}^{+\infty} a_n (x + \mathrm{i}y)^n$$ Show that $f$ is of class $\mathcal{C}^1$ on $D(0,R)$ and that its partial derivatives expand as power series on $D(0,R)$. What can we deduce about the function $f$?

A function $f$, defined on $D(0,R) \subset \mathbb{R}^2$ and with complex values, expands as a power series on $D(0,R)$ if there exists a complex sequence $(a_n)$ such that $$\forall (x,y) \in D(0,R), \quad f(x,y) = \sum_{n=0}^{+\infty} a_n (x + \mathrm{i}y)^n$$ We denote by $u$ and $v$ the real and imaginary parts of $f$, so that, for any $(x,y) \in D(0,R)$, $$u(x,y) \in \mathbb{R}, \quad v(x,y) \in \mathbb{R}, \quad f(x,y) = u(x,y) + \mathrm{i}v(x,y).$$ Show that $u$ and $v$ are harmonic functions on $D(0,R)$.

We say that a function $f$, defined on $D(0,R) \subset \mathbb{R}^2$ and with complex values, expands in a power series on $D(0,R)$ if there exists a complex sequence $(a_n)$ such that $$\forall (x,y) \in D(0,R), \quad f(x,y) = \sum_{n=0}^{+\infty} a_n (x + \mathrm{i} y)^n$$ Throughout this part, $f$ denotes a function that expands in a power series on $D(0,R)$.
Show that $f$ is of class $\mathcal{C}^1$ on $D(0,R)$ and that its partial derivatives expand in power series on $D(0,R)$. What can we deduce about the function $f$?

We say that a function $f$, defined on $D(0,R) \subset \mathbb{R}^2$ and with complex values, expands in a power series on $D(0,R)$ if there exists a complex sequence $(a_n)$ such that $$\forall (x,y) \in D(0,R), \quad f(x,y) = \sum_{n=0}^{+\infty} a_n (x + \mathrm{i} y)^n$$ We denote by $u$ and $v$ the real and imaginary parts of $f$, so that, for any $(x,y) \in D(0,R)$, $$u(x,y) \in \mathbb{R}, \quad v(x,y) \in \mathbb{R}, \quad f(x,y) = u(x,y) + \mathrm{i} v(x,y).$$ Show that $u$ and $v$ are harmonic functions on $D(0,R)$.

We denote $\mathcal { D }$ the open unit disk of $\mathbb { C } : \mathcal { D } = \{ z \in \mathbb { C } ; | z | < 1 \}$. The sequence $(a_n)$ is defined by $$\left\{ \begin{array} { l } a _ { 0 } = 1 \\ a _ { n } = \sum _ { k = 0 } ^ { n - 1 } \frac { ( - 1 ) ^ { n - k } } { ( n - k ) ! } H _ { k } \left( \frac { n + k } { 2 } - 1 \right) \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Show that, for all $z \in \mathcal { D }$, the power series $\sum _ { n \geqslant 0 } a _ { n } z ^ { n }$ converges.

We denote $\mathcal { D }$ the open unit disk of $\mathbb { C } : \mathcal { D } = \{ z \in \mathbb { C } ; | z | < 1 \}$. For $z \in \mathcal { D }$, we denote $\Phi _ { 0 } ( z ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } z ^ { n }$ and, subject to convergence, $$\Phi _ { p } ( z ) = \sum _ { n = 0 } ^ { + \infty } ( n + p ) ( n + p - 1 ) \cdots ( n + 1 ) a _ { n + p } z ^ { n }$$
Justify that, for all $p \in \mathbb { N } ^ { * }$ and all $\left. x \in \right] - 1,1 \left[ , \Phi _ { p } ( x ) = \varphi ^ { ( p ) } ( x ) \right.$ and that for all $p \in \mathbb { N } ^ { * }$ and all $z \in \mathcal { D }$, the power series $\sum _ { n = 0 } ^ { + \infty } ( n + p ) ( n + p - 1 ) \cdots ( n + 1 ) a _ { n + p } z ^ { n }$ converges.

We denote $\mathcal { D }$ the open unit disk of $\mathbb { C } : \mathcal { D } = \{ z \in \mathbb { C } ; | z | < 1 \}$. We define the function $\varphi : \mathbb { R } \rightarrow \mathbb { R }$ by $$\begin{cases} \varphi ( x ) = \exp \left( \frac { - x } { \sqrt { 1 - x } } \right) & \text { if } x < 1 \\ \varphi ( x ) = 0 & \text { if } x \geqslant 1 \end{cases}$$
Justify that, for all $p \in \mathbb { N }$, the function $\varphi ^ { ( p ) }$ is bounded on $] - 1,1 [$.

We denote $\mathcal { D }$ the open unit disk of $\mathbb { C } : \mathcal { D } = \{ z \in \mathbb { C } ; | z | < 1 \}$. For $z \in \mathcal { D }$, we denote $\Phi _ { p } ( z ) = \sum _ { n = 0 } ^ { + \infty } ( n + p ) ( n + p - 1 ) \cdots ( n + 1 ) a _ { n + p } z ^ { n }$. We admit that the function $\Phi _ { p }$ is bounded on $\mathcal { D }$.
Let $r$ be a real number in the interval $] 0,1 [$. Demonstrate for all integers $n \geqslant 1$ and $p \geqslant 1$, that $$( n + p ) ( n + p - 1 ) \cdots ( n + 1 ) a _ { n + p } r ^ { n } = \frac { 1 } { 2 \pi } \int _ { 0 } ^ { 2 \pi } \Phi _ { p } \left( r \mathrm { e } ^ { \mathrm { i } \theta } \right) \mathrm { e } ^ { - n \mathrm { i } \theta } \mathrm {~d} \theta$$

We consider a general balanced urn. For all real $x, u$ and $v$, we set $$H(x, u, v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$$ defined on $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2}$ for $\rho$ sufficiently small.
Prove that $H$ admits a first-order partial derivative with respect to $u$ on the domain $D_{\rho}$, obtained by term-by-term differentiation with respect to $u$ of the expression for $H$.

In the general model of a Pólya urn, we consider the balanced urn model for which $b = c = 0$, so $a = d$. Each time we draw a ball, we add $a$ balls of its color to the urn. The initial composition is $a_{0}$ white balls and $b_{0}$ black balls. The function $G$ is defined on $U = \{(x,u,v) \in \mathbb{R} \times \mathbb{R}_{+}^{*} \times \mathbb{R}_{+}^{*} ; axu^{a} < 1, axv^{a} < 1\}$ by $$G(x,u,v) = u^{a_{0}} v^{b_{0}} (1 - axu^{a})^{-a_{0}/a} (1 - axv^{a})^{-b_{0}/a}$$ We use the notation $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2}$.
Using the preliminary results, prove that there exists $\rho > 0$ such that $D_{\rho} \subset U$ and, for all $(x,u,v) \in D_{\rho}$, $$G(x,u,v) = \sum_{n=0}^{+\infty} Q_{n}(u,v) \frac{x^{n}}{n!}$$ where $Q_{n}$ is a polynomial function of two variables to be specified.

Let $(a_n)_n \in \mathbb{R}^{\mathbb{N}}$ be a sequence of reals such that the series $\sum (a_n)^2$ is convergent. Show that the radius of convergence of the power series $\sum a_n t^n$ is greater than or equal to 1.

Let $(a_n)_n \in \mathbb{R}^{\mathbb{N}}$ be a sequence of reals such that the series $\sum (a_n)^2$ is convergent. Show that the radius of convergence of the power series $\sum a_n t^n$ is greater than or equal to 1.