Imagine that we have a function $f:\mathbb{R} \rightarrow \mathbb{R}$.
We can find a power series for this function centered at 0: $\sum a_n x^n$ and we write $\sum a_n x^n=f(x)$
But we can also find a power series for this same function centered at 1, for example: $\sum a_n (x-1)^n$ and because this is a power series for the same function it would make sense to write $\sum a_n (x-1)^n=f(x)$.
We can do this for any $a \in \mathbb{R}$:$\sum a_n (x-a)^n$ and $\sum a_n (x-a)^n=f(x)$.
My question is: If we change the point where our power series is centered in, are we still getting a power series that converges to the same function or will we end up with $\sum a_n (x-a)^n=f(x-a)$?
In one hand it would make scene that $\sum a_n x^n \neq \sum a_n (x-1)^n$ but, on the other hand we say that this two series are power series of the same function, just centered at different points so it would make scene that $\sum a_n x^n = \sum a_n (x-1)^n = f(x)$
If you have a function $f(x)$, then $g(x) = f(x-1)$ is not the same as $f(x)$. We would say that $g(x)$ and $f(x)$ are the same "up to the symmetry $x\mapsto x-1$" which is a fancy way to say that $g(x)$ is obtained from $f(x)$ by shifting it to the right by $1$ unit.
Now, if $f(x)$ has a power series representation as $f(x) = \sum a_nx^n$ around $0$, then $g(x) = f(x-1) = \sum a_n(x-1)^n$ is still not the same as $f(x)$ by the discussion in the last paragraph. If we want to expand $f(x)$ about another point, say $x = 1$, then we do the following:
\begin{align*} f(x) = f(1 + (x-1)) &= \sum_{n=0}^\infty a_n(1 + (x-1))^n \\ &= \sum_{n=0}^\infty a_n\sum_{k=0}^n{n\choose k}(x-1)^k\\ &= \sum_{k=0}^\infty\underbrace{\Big(\sum_{n=k}^\infty a_n{n\choose k}\Big)}_{:= b_k}(x-1)^k\\ &:= \sum_{k=0}^\infty b_k(x-1)^k \end{align*} so we see that the coefficients $\{a_n\}_{n=0}^\infty$ of the power series will change to $\{b_k\}_{k=0}^\infty$, and in general $a_k\ne b_k$.