A function $f$ is defined by $$f(x) = \frac{1}{3} + \frac{2}{3^2}x + \frac{3}{3^3}x^2 + \cdots + \frac{n+1}{3^{n+1}}x^n + \cdots$$ for all $x$ in the interval of convergence of the given power series.
(a) Find the interval of convergence for this power series. Show the work that leads to your answer.
(b) Find $\displaystyle\lim_{x \rightarrow 0} \frac{f(x) - \frac{1}{3}}{x}$.
(c) Write the first three nonzero terms and the general term for an infinite series that represents $\displaystyle\int_0^1 f(x)\, dx$.
(d) Find the sum of the series determined in part (c).

The function $g$ has derivatives of all orders, and the Maclaurin series for $g$ is $$\sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{2n+3} = \frac{x}{3} - \frac{x^{3}}{5} + \frac{x^{5}}{7} - \cdots$$
(a) Using the ratio test, determine the interval of convergence of the Maclaurin series for $g$.
(b) The Maclaurin series for $g$ evaluated at $x = \frac{1}{2}$ is an alternating series whose terms decrease in absolute value to 0. The approximation for $g\left(\frac{1}{2}\right)$ using the first two nonzero terms of this series is $\frac{17}{120}$. Show that this approximation differs from $g\left(\frac{1}{2}\right)$ by less than $\frac{1}{200}$.
(c) Write the first three nonzero terms and the general term of the Maclaurin series for $g'(x)$.

What is the radius of convergence of the series $\sum _ { n = 0 } ^ { \infty } \frac { ( x - 4 ) ^ { 2 n } } { 3 ^ { n } }$ ?
(A) $2 \sqrt { 3 }$
(B) 3
(C) $\sqrt { 3 }$
(D) $\frac { \sqrt { 3 } } { 2 }$
(E) 0

The power series $\sum _ { n = 0 } ^ { \infty } a _ { n } ( x - 3 ) ^ { n }$ converges at $x = 5$. Which of the following must be true?
(A) The series diverges at $x = 0$.
(B) The series diverges at $x = 1$.
(C) The series converges at $x = 1$.
(D) The series converges at $x = 2$.
(E) The series converges at $x = 6$.

The Taylor series for a function $f$ about $x = 1$ is given by $\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { 2 ^ { n } } { n } ( x - 1 ) ^ { n }$ and converges to $f ( x )$ for $| x - 1 | < R$, where $R$ is the radius of convergence of the Taylor series.
(a) Find the value of $R$.
(b) Find the first three nonzero terms and the general term of the Taylor series for $f ^ { \prime }$, the derivative of $f$, about $x = 1$.
(c) The Taylor series for $f ^ { \prime }$ about $x = 1$, found in part (b), is a geometric series. Find the function $f ^ { \prime }$ to which the series converges for $| x - 1 | < R$. Use this function to determine $f$ for $| x - 1 | < R$.

The Maclaurin series for a function $f$ is given by $$\sum _ { n = 1 } ^ { \infty } \frac { ( - 3 ) ^ { n - 1 } } { n } x ^ { n } = x - \frac { 3 } { 2 } x ^ { 2 } + 3 x ^ { 3 } - \cdots + \frac { ( - 3 ) ^ { n - 1 } } { n } x ^ { n } + \cdots$$ and converges to $f ( x )$ for $| x | < R$, where $R$ is the radius of convergence of the Maclaurin series.
(a) Use the ratio test to find $R$.
(b) Write the first four nonzero terms of the Maclaurin series for $f ^ { \prime }$, the derivative of $f$. Express $f ^ { \prime }$ as a rational function for $| x | < R$.
(c) Write the first four nonzero terms of the Maclaurin series for $e ^ { x }$. Use the Maclaurin series for $e ^ { x }$ to write the third-degree Taylor polynomial for $g ( x ) = e ^ { x } f ( x )$ about $x = 0$.

The function $f$ has a Taylor series about $x = 1$ that converges to $f ( x )$ for all $x$ in the interval of convergence. It is known that $f ( 1 ) = 1 , f ^ { \prime } ( 1 ) = - \frac { 1 } { 2 }$, and the $n$th derivative of $f$ at $x = 1$ is given by $f ^ { ( n ) } ( 1 ) = ( - 1 ) ^ { n } \frac { ( n - 1 ) ! } { 2 ^ { n } }$ for $n \geq 2$.
(a) Write the first four nonzero terms and the general term of the Taylor series for $f$ about $x = 1$.
(b) The Taylor series for $f$ about $x = 1$ has a radius of convergence of 2. Find the interval of convergence. Show the work that leads to your answer.
(c) The Taylor series for $f$ about $x = 1$ can be used to represent $f ( 1.2 )$ as an alternating series. Use the first three nonzero terms of the alternating series to approximate $f ( 1.2 )$.
(d) Show that the approximation found in part (c) is within 0.001 of the exact value of $f ( 1.2 )$.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{t}{e^t - 1} - 1 + \dfrac{t}{2}$ for all $t \in \mathbb{R}^{+*}$.
Show that $f$ extends by continuity at 0.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{t}{e^t - 1} - 1 + \dfrac{t}{2}$ for all $t \in \mathbb{R}^{+*}$ (extended by continuity at 0).
Using a series expansion, show that for all $x > 0$, we have $$Lf(x) = \frac{1}{2x^2} - \frac{1}{x} + \sum_{n=1}^{+\infty} \frac{1}{(n+x)^2}.$$

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence. We assume that the associated power series $\sum a _ { n } x ^ { n }$ has radius of convergence $R _ { a } = 1$ and that the sum $f$ of this series satisfies $f ( x ) \sim \frac { 1 } { 1 - x }$ when $x \rightarrow 1, x < 1$.
Determine a real sequence $\left( b _ { n } \right) _ { n \geqslant 0 }$ such that $$\forall x \in ] - 1,1 [ , \quad \frac { 1 } { 1 - x ^ { 2 } } = \sum _ { n = 0 } ^ { + \infty } b _ { n } x ^ { n }$$

Let $\left( a _ { n } \right) _ { n \geqslant 0 }$ be a real sequence. We assume that the associated power series $\sum a _ { n } x ^ { n }$ has radius of convergence $R _ { a } = 1$ and that the sum $f$ of this series satisfies $f ( x ) \sim \frac { 1 } { 1 - x }$ when $x \rightarrow 1, x < 1$.
Deduce an example of a sequence $\left( a _ { n } \right) _ { n \geqslant 0 }$ satisfying hypothesis II.1 ($f(x) \sim \frac{1}{1-x}$ as $x \to 1^-$) but not converging to 1.

Give the power series expansion of the function $t \mapsto \frac { 1 } { ( 1 - t ) ^ { 2 } }$ as well as its radius of convergence. Specify whether the series converges at the endpoints of the interval of convergence.

Show that for real $x$,
$$\varphi _ { n } ( x ) = \sum _ { k = 0 } ^ { + \infty } \frac { x ^ { k } } { k ! } I _ { n , k } \quad \text { with } \quad I _ { n , k } = \frac { 1 } { 2 \pi } \int _ { - \pi } ^ { \pi } i ^ { k } e ^ { - i n t } ( \sin t ) ^ { k } \mathrm {~d} t$$

Deduce the power series development, for $n \geqslant 0$ and $x \in \mathbb { R }$ :
$$\varphi _ { n } ( x ) = \sum _ { p = 0 } ^ { + \infty } \frac { ( - 1 ) ^ { p } } { p ! ( n + p ) ! } \left( \frac { x } { 2 } \right) ^ { n + 2 p }$$
Specify the radius of convergence.

Let $\theta \in \mathbb { R }$.
Determine the radius of convergence of the power series $\sum \frac { e ^ { \mathrm { i n } \theta } } { n } x ^ { n }$.

Give without proof the power series expansions of the functions $t \mapsto \ln ( 1 - t )$ and $t \mapsto \frac { 1 } { 1 - t }$ as well as their radius of convergence.

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k } = 1 + \frac { 1 } { 2 } + \cdots + \frac { 1 } { n }$.
Deduce that the function $$t \mapsto - \frac { \ln ( 1 - t ) } { 1 - t }$$ is expandable as a power series on $] - 1,1 [$ and specify its power series expansion using the real numbers $H _ { n }$.

We denote $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$. We denote $g$ the function defined on $[ - 1 , + \infty [$ by $$g ( x ) = \sum _ { n = 2 } ^ { + \infty } \left( \frac { 1 } { n } - \frac { 1 } { n + x } \right)$$
Show that for every integer $n$ and for every $x$ in $] - 1,1 [$ $$\left| g ( x ) - \sum _ { k = 0 } ^ { n } \frac { g ^ { ( k ) } ( 0 ) } { k ! } x ^ { k } \right| \leqslant \zeta ( 2 ) | x | ^ { n + 1 }$$
Show that $g$ is expandable as a power series on $] - 1,1 [$.

We denote $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$. We define $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$ on $\mathbb{R}^{+*}$. We have shown that for every real $x > -1$, $\psi ( 1 + x ) = \psi ( 1 ) + \sum _ { n = 1 } ^ { + \infty } \left( \frac { 1 } { n } - \frac { 1 } { n + x } \right)$.
Prove that for every $x$ in $] - 1,1 [$, $$\psi ( 1 + x ) = \psi ( 1 ) + \sum _ { n = 1 } ^ { + \infty } ( - 1 ) ^ { n + 1 } \zeta ( n + 1 ) x ^ { n }$$

We recall that $\phi$ is defined on $] - 1 , + \infty [$ by $\phi ( s ) = s - \ln ( 1 + s )$.
Show that if $s \in ] - 1,1 [$, $$\phi ( s ) = s ^ { 2 } \sum _ { k = 0 } ^ { \infty } \frac { ( - 1 ) ^ { k } } { k + 2 } s ^ { k }$$
We admit the existence of two power series $\sum _ { k \geqslant 1 } b _ { k } q ^ { k }$ and $\sum _ { k \geqslant 1 } c _ { k } q ^ { k }$, in the variable $q$, and of strictly positive radius of convergence, where $b _ { 1 } > 0$ and $c _ { 1 } < 0$, and such that we have, for $q$ in a neighborhood of 0 in $[ 0 , + \infty [$, $$\phi \left( \sum _ { k = 1 } ^ { \infty } b _ { k } q ^ { k } \right) = \phi \left( \sum _ { k = 1 } ^ { \infty } c _ { k } q ^ { k } \right) = q ^ { 2 } .$$

We set for every integer $n \geqslant 0$, $$B _ { n } = \sum _ { k = 0 } ^ { n } S ( n , k )$$ where $S(n,k)$ denotes the number of partitions of $\llbracket 1, n \rrbracket$ into $k$ parts. The sequence $\left( \frac { B _ { n } } { n ! } \right) _ { n \in \mathbb { N } }$ is bounded by 1.
Deduce a lower bound for the radius of convergence $R$ of the power series $\sum _ { n \geqslant 0 } \frac { B _ { n } } { n ! } z ^ { n }$.

We consider the power series in the real variable $x$ given by $\sum_{k \in \mathbb{N}^{*}} (-1)^{k} \zeta(k+1) x^{k}$.
Determine the radius of convergence $R$ of this power series. Is there convergence at $x = \pm R$?

Let $f$ be the function defined by $$f(x) = \sum_{n=1}^{+\infty} \left(\frac{1}{n+x} - \frac{1}{n}\right)$$ Using the result of Q18, deduce that $f$ is expandable as a power series on $]-1,1[$ and that $$\forall x \in {]-1,1[}, \quad f(x) = \sum_{k=1}^{+\infty} (-1)^{k} \zeta(k+1) x^{k}$$

We consider the Hilbert polynomials $$\left\{ \begin{array} { l } H _ { 0 } ( X ) = 1 \\ H _ { n } ( X ) = \frac { 1 } { n ! } \prod _ { k = 0 } ^ { n - 1 } ( X - k ) = \frac { X ( X - 1 ) \cdots ( X - n + 1 ) } { n ! } \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Demonstrate that, for all $x \in ] - 1,1 [$ and all $q \in \mathbb { N } ^ { * }$, we have $$( 1 - x ) ^ { - q / 2 } = \sum _ { p = 0 } ^ { + \infty } H _ { p } \left( \frac { q } { 2 } + p - 1 \right) x ^ { p }$$

Using the result of Q6, deduce the lower bound $R \geqslant \pi/2$ for the radius of convergence $R$ of the power series $\sum_{n \in \mathbb{N}} \frac{\alpha_n}{n!} x^n$.