<s>[INST] Estimating Residues in McLauren's Polynomial Can someone please help me figure out why it is enough to take $n\geq 9 $ in order to get that:
$$\displaystyle  \frac{ 9 \cdot 2^{n+1} } {(n+1)!} < \frac{1}{100}\;\;?$$
Thanks in advance
 [/INST] First note that $a_n = \dfrac{2^{n+1} \cdot 9}{(n+1)!}$ is a monotone decreasing sequence for $n \geq 1$. Hence, if $a_n < \epsilon$, then $a_m < \epsilon$ for all $m \geq n$.
$$a_n = \dfrac{2^{n+1} \cdot 9}{(n+1)!} = \dfrac{3 \cdot 2^{n-2} \cdot 4!}{(n+1)!}$$
$$a_8 = \dfrac{3 \cdot 2^6 \cdot 4!}{9!} = \dfrac{3 \cdot 2^6}{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5} = \dfrac{2^2}{9 \cdot 7 \cdot 5} = \dfrac4{315} > \dfrac4{400} = \dfrac1{100}$$
$$a_9 = \dfrac{3 \cdot 2^7 \cdot 4!}{10!} = \dfrac{3 \cdot 2^7}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5} = \dfrac{2^2}{5 \cdot 9 \cdot 7 \cdot 5} = \dfrac4{1575} < \dfrac4{400} = \dfrac1{100}$$
Since $a_9 < \dfrac1{100}$ and $a_n$ is a decreasing sequence, we have that $a_n < \dfrac1{100}$ for all $n \geq 9$.
</s>