In this question, we return to the case of II.B.3 (i.e., $\alpha = 3$, $p = 3$, $n = 100$). Knowing that $6 ! a _ { 6 } = \frac { 1 } { 42 }$, recover that the error $\left| S ( 3 ) - \widetilde { S } _ { 100,4 } ( 3 ) \right|$ is bounded by an expression of order $10 ^ { - 17 }$.

We approximate $\varphi _ { n } ( x )$ using partial sums
$$S _ { m } = \sum _ { p = 0 } ^ { m } ( - 1 ) ^ { p } a _ { p } \quad \text { with } \quad m \in \mathbb { N } \quad \text { and } \quad a _ { p } = \frac { 1 } { p ! ( n + p ) ! } \left( \frac { x } { 2 } \right) ^ { n + 2 p }$$
Assume $N > p _ { 0 }$. Bound $\left| R _ { N } \right|$ as a function of $(N, n, x)$ with
$$R _ { N } = \sum _ { p = N + 1 } ^ { + \infty } ( - 1 ) ^ { p } a _ { p }$$
Deduce, for fixed $\varepsilon > 0$, a sufficient condition on $N$ for $\left| \varphi _ { n } ( x ) - S _ { N } \right| < \varepsilon$.

We consider the function $F : ] 0 , + \infty [ \rightarrow \mathbb { R }$ defined by $F ( x ) = \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - 1 } d t$.
For $N \in \mathbb { N } ^ { * }$ and $x > 0$, we set $$\begin{aligned} r _ { N } ( x ) & = ( - 1 ) ^ { N } N ! x ^ { N + 1 } e ^ { - 1 / x } \\ S _ { N } ( x ) & = \sum _ { k = 1 } ^ { N } ( - 1 ) ^ { k - 1 } ( k - 1 ) ! x ^ { k } e ^ { - 1 / x } \\ R _ { N } ( x ) & = ( - 1 ) ^ { N } N ! x ^ { N } \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - ( N + 1 ) } d t \end{aligned}$$ The relative error is $E _ { N } ( x ) = \left| \frac { R _ { N } ( x ) } { F ( x ) } \right|$.
Show that, if $N$ is even: $N = 2 M$ with $M \geqslant 1$, and if $0 < x \leqslant 1 / N$, we have $S _ { N } ( x ) \geqslant 0$ and $$E _ { N } ( x ) \leqslant \frac { N ! x ^ { N + 1 } } { \sum _ { \ell = 0 } ^ { M - 1 } ( 1 - ( 2 \ell + 1 ) x ) ( 2 \ell ) ! x ^ { 2 \ell + 1 } }$$

We consider the function $F : ] 0 , + \infty [ \rightarrow \mathbb { R }$ defined by $F ( x ) = \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - 1 } d t$.
For $N \in \mathbb { N } ^ { * }$ and $x > 0$, we set $$\begin{aligned} S _ { N } ( x ) & = \sum _ { k = 1 } ^ { N } ( - 1 ) ^ { k - 1 } ( k - 1 ) ! x ^ { k } e ^ { - 1 / x } \\ R _ { N } ( x ) & = ( - 1 ) ^ { N } N ! x ^ { N } \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - ( N + 1 ) } d t \end{aligned}$$ The relative error is $E _ { N } ( x ) = \left| \frac { R _ { N } ( x ) } { F ( x ) } \right|$.
Verify that $E _ { 4 } \left( \frac { 1 } { 10 } \right) \leqslant 3.10 ^ { - 3 }$.

Using the result of Q5, deduce that for all $x \in \mathcal{D}_{\zeta}$, $$1 + \frac{1}{(x-1)2^{x-1}} \leqslant \zeta(x) \leqslant 1 + \frac{1}{x-1}$$

For every $s > 1$, let $\zeta(s) = \sum_{n=1}^{+\infty} \frac{1}{n^s}$. Bound $\sum_{n=2}^{+\infty} \frac{1}{n^s}$ by two integrals and deduce $\lim_{s \rightarrow +\infty} \zeta(s) = 1$.

We set $C = \exp\left(\frac{I}{2\pi}\right)$ with $I = 4\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}$. The calculator gives: $$\begin{aligned} \exp\left(\frac{2}{\pi}\sum_{k=0}^5 \frac{(-1)^k}{(2k+1)^2}\right) &\approx 1{,}78774486868 \\ \exp\left(\frac{2}{\pi}\sum_{k=0}^6 \frac{(-1)^k}{(2k+1)^2}\right) &\approx 1{,}79449196958 \end{aligned}$$ Can we deduce the rounding of $C$ to $10^{-2}$ precision? If yes, give the value of this rounding. In any case, justify the answer properly.

Show that $$\int _ { 1 } ^ { + \infty } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t = O \left( \frac { 1 } { n 2 ^ { n } } \right) .$$ One may lower bound $1 + t ^ { 2 }$ by a polynomial of degree 1.

Find the integer part of $S = \displaystyle\sum_{k=2}^{9999} \dfrac{1}{\sqrt{k}}$.

For any $n \geq 5$, the value of $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n - 1}$ lies between
(A) 0 and $\frac{n}{2}$
(B) $\frac{n}{2}$ and $n$
(C) $n$ and $2n$
(D) none of the above.

For any $n \geq 5$, the value of $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n - 1}$ lies between
(A) 0 and $\frac{n}{2}$
(B) $\frac{n}{2}$ and $n$
(C) $n$ and $2n$
(D) none of the above

For any $n \geq 5$, the value of $1 + \frac { 1 } { 2 } + \frac { 1 } { 3 } + \cdots + \frac { 1 } { 2 ^ { n } - 1 }$ lies between
(A) 0 and $\frac { n } { 2 }$
(B) $\frac { n } { 2 }$ and $n$
(C) $n$ and $2 n$
(D) none of the above

The integral part of $\sum _ { n = 2 } ^ { 9999 } \frac { 1 } { \sqrt { n } }$ equals
(A) 196
(B) 197
(C) 198
(D) 199 .

Let $$S = \frac{1}{\sqrt{10000}} + \frac{1}{\sqrt{10001}} + \cdots + \frac{1}{\sqrt{160000}}$$ Then the largest positive integer not exceeding $S$ is
(A) 200
(B) 400
(C) 600
(D) 800

For any $n \geq 5$, the value of $1 + \frac { 1 } { 2 } + \frac { 1 } { 3 } + \cdots + \frac { 1 } { 2 ^ { n } - 1 }$ lies between
(a) 0 and $n / 2$.
(B) $n / 2$ and $n$.
(C) $n$ and $2 n$.
(D) none of the above.