$\sum_{n=1}^{\infty}(a_{n}-a_{n-1})x^n$ and $\sum_{n=0}^{\infty}a_nx^n$ has the same convergence radius Proof:
Suppose power series $\sum_{n=0}^{\infty}a_n x^n $, if series ${a_n}$ is divergent. Then $\sum_{n=1}^{\infty}(a_{n}-a_{n-1})x^n$ and $\sum_{n=0}^{\infty}a_nx^n$ has the same convergence radius.

I only can see that when $x=1$ these are both divergent. Thank you.
EDIT: Solution is WRONG. I would delete if I could.
Let $R$ be the radius of convergence of $\sum_{n=0}^\infty a_n x^n$. That is, $\sum_{n=0}^\infty a_n x^n$ converges when $|x| &lt; R$, but diverges when $|x| &gt; R$. We need to show $\sum_{n=1}^\infty (a_n - a_{n-1}) x^n$ converges when $|x| &lt; R$, and diverges when $|x| &gt; R$.
If $|x| &lt; R$, then
$$\sum_{n=1}^\infty (a_n - a_{n-1}) x^n = \sum_{n=1}^\infty a_n x^n - x\sum_{n=0}^\infty a_n x^n,$$
i.e. a combination of two convergent series, making the series convergent.
Suppose $|x| &gt; R$. Without loss of generality, we may assume $x \neq 1$; if $x = 1$ is possible, then $R &lt; 1$, and the $x = 1$ case is covered by $x = \frac{1 + R}{2}$.
For the sake of contradiction, suppose $\sum_{n=1}^\infty (a_n - a_{n-1}) x^n = L$. We can similarly assume that this sum is absolutely convergent, since we can replace $x$ with $\frac{R + x}{2}$, taking $x$ off the boundary of the interval of convergence. Then
\begin{align*}
\frac{L}{1 - x} &amp;= \sum_{m=0}^\infty x^mL \\
&amp;= \sum_{m=0}^\infty x^m\sum_{n=1}^\infty (a_n - a_{n-1}) x^n \\
&amp;= x\sum_{n=0}^\infty \sum_{m=0}^\infty (a_{n+1} - a_n) x^{m+n}.
\end{align*}
I wish to express $\sum_{n=0}^\infty \sum_{m=0}^\infty (a_{n+1} - a_n) x^{m+n}$ as a simple power series, collecting each power of $x$ as a single term. Note that the $x^k$ term occurs when $n + m = k$, giving us:
\begin{align*}
\frac{L}{1 - x} &amp;= x\sum_{k=0}^\infty\sum_{l=0}^k (a_{l+1} - a_l) x^k \\
&amp;= x\sum_{k=0}^\infty(a_{k+1} - a_0)x^k \\
&amp;= \sum_{k=0}^\infty a_{k+1} x^{k+1} - a_0 x \sum_{k=0}^\infty x^k \\
&amp;= \sum_{k=0}^\infty a_k x^k - a_0 - a_0 x \sum_{k=0}^\infty x^k \\
&amp;= \sum_{k=0}^\infty a_k x^k - a_0 \sum_{k=0}^\infty x^k \\
&amp;= \sum_{k=0}^\infty a_k x^k - \frac{a_0}{1 - x} \\
\frac{L + a_0}{1 - x} &amp;= \sum_{k=0}^\infty a_k x^k.
\end{align*}
That is, $\sum_{k=0}^\infty a_k x^k$ is convergent too, contradicting $|x| &gt; R$.
I think maybe I found the solution:
When $x=1$, the two are both divergent, so since they are power series, when $x\ge 1$, it is divergent.
When $x&lt; 1$, by summation by part (Abel transformation),
$$\sum_{n=1}^{N}(a_n-a_{n-1})x^{n}\\=(x^na_n-xa_0)-(1-\frac{1}{x})\sum_{n=1}^{N-1}a_nx^n,$$ since $x^n$ is convergence when $x&lt;1$, the two series has the same convergence and divergence.