The Maclaurin series for $\ln\left(\dfrac{1}{1-x}\right)$ is $\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n}$ with interval of convergence $-1 \leq x < 1$.
(a) Find the Maclaurin series for $\ln\left(\dfrac{1}{1+3x}\right)$ and determine the interval of convergence.
(b) Find the value of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$.
(c) Give a value of $p$ such that $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^p}$ converges, but $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2p}}$ diverges. Give reasons why your value of $p$ is correct.
(d) Give a value of $p$ such that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges, but $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2p}}$ converges. Give reasons why your value of $p$ is correct.

Let $f$ be the function given by $f(x) = e^{-x^2}$.
(a) Write the first four nonzero terms and the general term of the Taylor series for $f$ about $x = 0$.
(b) Use your answer to part (a) to find $\lim_{x \to 0} \frac{1 - x^2 - f(x)}{x^4}$.
(c) Write the first four nonzero terms of the Taylor series for $\int_{0}^{x} e^{-t^2}\, dt$ about $x = 0$. Use the first two terms of your answer to estimate $\int_{0}^{1/2} e^{-t^2}\, dt$.
(d) Explain why the estimate found in part (c) differs from the actual value of $\int_{0}^{1/2} e^{-t^2}\, dt$ by less than $\frac{1}{200}$.

The Maclaurin series for $e^{x}$ is $e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \cdots + \frac{x^{n}}{n!} + \cdots$. The continuous function $f$ is defined by $f(x) = \frac{e^{(x-1)^{2}} - 1}{(x-1)^{2}}$ for $x \neq 1$ and $f(1) = 1$. The function $f$ has derivatives of all orders at $x = 1$.
(a) Write the first four nonzero terms and the general term of the Taylor series for $e^{(x-1)^{2}}$ about $x = 1$.
(b) Use the Taylor series found in part (a) to write the first four nonzero terms and the general term of the Taylor series for $f$ about $x = 1$.
(c) Use the ratio test to find the interval of convergence for the Taylor series found in part (b).
(d) Use the Taylor series for $f$ about $x = 1$ to determine whether the graph of $f$ has any points of inflection.

The Maclaurin series for $\ln ( 1 + x )$ is given by
$$x - \frac { x ^ { 2 } } { 2 } + \frac { x ^ { 3 } } { 3 } - \frac { x ^ { 4 } } { 4 } + \cdots + ( - 1 ) ^ { n + 1 } \frac { x ^ { n } } { n } + \cdots$$
On its interval of convergence, this series converges to $\ln ( 1 + x )$. Let $f$ be the function defined by
$$f ( x ) = x \ln \left( 1 + \frac { x } { 3 } \right)$$
(a) Write the first four nonzero terms and the general term of the Maclaurin series for $f$.
(b) Determine the interval of convergence of the Maclaurin series for $f$. Show the work that leads to your answer.
(c) Let $P _ { 4 } ( x )$ be the fourth-degree Taylor polynomial for $f$ about $x = 0$. Use the alternating series error bound to find an upper bound for $\left| P _ { 4 } ( 2 ) - f ( 2 ) \right|$.

Find the Taylor series expansion of $f ( x ) = e ^ { x } \sin ( x )$ centered at $x = \frac { \pi } { 4 }$.

We recall that the hyperbolic cosine function, which we denote cosh, is defined, for every real $t$, by $$\cosh(t)=\frac{\mathrm{e}^{t}+\mathrm{e}^{-t}}{2}$$
a) Give the power series expansion of the hyperbolic cosine function and that of the function defined on $\mathbb{R}$ by $t \mapsto \mathrm{e}^{t^{2}/2}$. We will give the radius of convergence of these two power series.
b) Deduce that $\forall t \in \mathbb{R}, \cosh(t) \leqslant \mathrm{e}^{t^{2}/2}$.

Deduce the existence of a real $\nu_{\sigma}$ such that, for any real $\xi$ and any real $t > 0$, $$\hat{f}(t, \xi) = \nu_{\sigma} \exp\left(-2\pi^{2}\left(\sigma^{2}+2t\right) \xi^{2}\right)$$

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$.
Show that: $$\forall x \in ]-2\pi, 2\pi[ \backslash \{0\}, \quad \frac{x}{2} \operatorname{cotan}\left(\frac{x}{2}\right) = 1 - \sum_{k=1}^{+\infty} \frac{\zeta(2k)}{2^{2k-1}\pi^{2k}} x^{2k}.$$

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$.
Deduce: $$\forall x \in ]-2\pi, 2\pi[ \backslash \{0\}, \quad \frac{ix}{e^{ix}-1} = 1 - \frac{ix}{2} - \sum_{k=1}^{+\infty} \frac{\zeta(2k)}{2^{2k-1}\pi^{2k}} \cdot x^{2k}.$$

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$. Using the identity $\pi x \operatorname{cotan}(\pi x) = 1 + 2\sum_{n=1}^{+\infty} \frac{x^2}{x^2 - n^2}$, show that: $$\forall x \in \left]-2\pi, 2\pi\right[ \backslash \{0\}, \quad \frac{x}{2}\operatorname{cotan}\left(\frac{x}{2}\right) = 1 - \sum_{k=1}^{+\infty} \frac{\zeta(2k)}{2^{2k-1}\pi^{2k}} x^{2k}$$

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$. Using the result $$\forall x \in \left]-2\pi, 2\pi\right[ \backslash \{0\}, \quad \frac{x}{2}\operatorname{cotan}\left(\frac{x}{2}\right) = 1 - \sum_{k=1}^{+\infty} \frac{\zeta(2k)}{2^{2k-1}\pi^{2k}} x^{2k}$$ deduce: $$\forall x \in \left]-2\pi, 2\pi\right[ \backslash \{0\}, \quad \frac{ix}{e^{ix}-1} = 1 - \frac{ix}{2} - \sum_{k=1}^{+\infty} \frac{\zeta(2k)}{2^{2k-1}\pi^{2k}} \cdot x^{2k}$$

Using the result of Q11, deduce that for all $\zeta \in \mathbb{U}$ and all $p \in \mathbb{Z}$, $$\frac{\zeta^{p}}{\mathrm{e}^{\zeta} - 1} = \sum_{j=0}^{+\infty} (-1)^{j} \zeta^{j+p-1} \beta(\zeta)^{j}$$ where $\beta \in \mathcal{E}$ and $|\beta(\zeta)| \leqslant C < 1$ for all $\zeta \in \mathbb{U}$.

Recall that $x$ is a fixed element of $]0;1[$. Finally deduce that:
$$\forall y \in ] 0 ; \pi \left[ , \quad \sum _ { n = 1 } ^ { + \infty } \frac { 2 ( - 1 ) ^ { n } y \sin ( y ) } { y ^ { 2 } - n ^ { 2 } \pi ^ { 2 } } = 1 - \frac { \sin ( y ) } { y } . \right.$$