Change in Interval of convergence if center of convergence changes So I have to find a power series that is centered at $-2^{1/2}$
If I choose to use the power series expansion for $e^x$ which converges for all $x$, and change $x$ to $x + 2^{1/2}$ does the interval of convergence remain the same?
 A: If the radius of convergence of a power series is positive and finite, it is the distance from the center of the series expansion to the nearest singularity or branchpoint. If the center changes, the distance to the nearest singularity or branchpoint may change and so the radius of convergence will also change.
Considering for example the geometric series expansion at $x=0$, we obtain a power series with radius of convergence equal to $1$, since we have a singularity, a pole of order $1$ at $x=1$.
\begin{align*}
\frac{1}{1-x}=\sum_{n=0}^\infty x^n\qquad\qquad|x|<1
\end{align*}
If we expand the geometric series at $x=\frac{1}{2}$ we obtain a power series with radius of convergence equal to $\frac{1}{2}$, since then the distance to the nearest singularity $1$ from the center $x=\frac{1}{2}$ is equal to $\frac{1}{2}$.
\begin{align*}
\frac{1}{1-x}=\frac{2}{1-2\left(x-\frac{1}{2}\right)}=\sum_{n=0}^\infty2^{n+1}\left(x-\frac{1}{2}\right)^n\qquad\qquad\left|x-\frac{1}{2}\right|<\frac{1}{2}
\end{align*}

The situation is different, when we consider an entire function which is convergent in the whole complex plane. Since the radius of convergence is $\infty$ in this case, it will not change if we change the center. The exponential function is an entire function and so
  \begin{align*}
e^x&=\sum_{n=0}^\infty\frac{x^n}{n!}\qquad\qquad\qquad\qquad x\in \mathbb{C}\\
e^x=e^{x_0}e^{x-x_0}&=e^{x_0}\sum_{n=0}^\infty \frac{\left(x-x_0\right)^n}{n!}\qquad\qquad x\in\mathbb{C}
\end{align*}

