$$ f \left( x \right) := \left( 1+x \right) ^{d} ~~ \leftarrow~~ d \in\mathbb{R} $$
I came across the problem of Maclaurin expansion.
$$ f \left( x \right) = \sum_{ i=0 }^{ \infty } \frac{ f^{\left( i \right) } \left( 0 \right) \cdot x ^{i} }{ i ! } $$
$$ f ^{\left( 0 \right) }\left( x \right) =\left( 1+x \right) ^{d} $$
$$ f ^{\left( 1\right) }\left( x \right)=\left( d \right) \left( 1+x \right) ^{d-1} $$
$$ f ^{\left( 2\right) }\left( x \right)=\left( d \right) \left( d-1 \right) \left( 1+x \right) ^{d-2} $$
$$ f ^{\left( 3\right) }\left( x \right)=\left( d \right) \left( d-1 \right) \left( d-2 \right) \left( 1+x \right) ^{d-3} $$
$$ \therefore ~~ f^{\left( i \right) }\left( x \right) = \left\{ \prod_{ j=0 }^{i-1 }\left(d-j \right) \right\} \cdot \left( 1+x \right) ^{d-i} $$
For instance , imagine that we set $~ d=-\frac{ 3 }{ 2 } ~$
$$ f^{\left( i \right) }\left( x \right) = \left\{ \prod_{ j=0 }^{i-1 }\left(\left( -\frac{ 3 }{ 2 } \right) -j \right) \right\} \cdot \left( 1+x \right) ^{d-i} $$
$$ = \left\{ \prod_{ j=0 }^{ i-1 } \left( -\frac{ 3 }{ 2 } -j \right) \right\} \cdot \left( 1+x \right) ^{d-i} $$
$$ = \left\{ \prod_{ j=0 }^{ i-1 } \left( \left( -1 \right) \left( \frac{ 3 }{ 2 } +j\right) \right) \right\} \cdot \left( 1+x \right) ^{d-i} $$
$$ = \left( -1 \right)^{i} \left\{ \prod_{ j=0 }^{ i-1 } \left( \frac{ 3 }{ 2 } +j\right) \right\} \cdot \left( 1+x \right) ^{d-i} $$
$$ = f^{\left( i \right) }\left( x \right) = \left( -1 \right)^{i} \left\{ \prod_{ j=0 }^{ i-1 } \left( \frac{ 3 }{ 2 } +j\right) \right\} \cdot \left( 1+x \right) ^{d-i}$$
Imagine we set $~ x=0 ~$
$$ f^{\left( i \right) }\left( 0 \right) = \left( -1 \right)^{i} \left\{ \prod_{ j=0 }^{ i-1 } \left( \frac{ 3 }{ 2 } +j\right) \right\} \cdot \left( 1+0 \right) ^{d-i}$$
$$ = \left( -1 \right)^{i} \left\{ \underbrace{\color{blue}{\prod_{ j=0 }^{ i-1 } \left( \frac{ 3 }{ 2 } +j\right)} }_{ \color{black}{\text{This value}} ~ \propto ~i } \right\} ~~ \leftarrow~~ \text{This value diverges to infinity} $$
Nextly we consider each term of Maclaurin series.
$$ \frac{ f^{\left( i \right) } \left( 0 \right) \cdot x ^{i} }{ i ! } $$
I've been known that as $~ x \approx 0 ~$ is held , we can assume the below rightmost formula.
$$ f \left( x \right) = \sum_{ i=0 }^{ \infty } \frac{ f^{\left( i \right) } \left( 0 \right) \cdot x ^{i} }{ i !} \approx \sum_{ i=0 }^{ 1 } \frac{ f^{\left( i \right) } \left( 0 \right) \cdot x ^{i} }{ i! } $$
One of the reason(s) that the above rightmost term is held is that $~ x^{\text{index greater than 1} } \approx 0 ~$ is held , can be said . But still I am concerning the term which I colored with blue .
$$ \frac{ f^{\left( i \right) } \left( 0 \right) \cdot x ^{i} }{ i ! }= \frac{ 1 }{ i! } \cdot \left( \color{red}{ f^{\left( i \right) } \left( 0 \right)}\right) \cdot \left( \color{green}{ x ^{i}} \right) ~~ \leftarrow~~ ~~ \text{Red one diverges to infinity and green one converges to zero}~~ $$
Why this term can be said to converges to zero?
But it is said that $~ f \left( x \right) = \left( 1+x \right) ^{d} ~$ can be represented with 2 terms as $~ x \approx 0 ~$
What is going on?