<s>[INST] Radius of convergence of power series $\sum_{n=0}^{\infty} \frac{(2x-5)^n}{n^2}$ Consider the below example: 
$$\sum_{n=0}^{\infty}\frac{(2x-5)^n}{n^2},\qquad c_n=\frac{2^n}{n^2},\qquad R=2^{-1}=\frac{1}{2}.$$
$$\begin{align*}
\lim_{n\to\infty}\frac{c_{n+1}}{c_n} &= \lim_{n\to\infty}\frac{2^{n+1}/(n+1)^2}{2^n/n^2} \\
&=\lim_{n\to\infty}\frac{2n^2}{(n+1)^2}\\
&= \lim_{n\to\infty}\frac{2}{(1+\frac{1}{n})^2}\\
&= 2.
\end{align*}$$
How come the expression 
$$\sum_{n=0}^{\infty} \frac{(2x-5)^n}{n^2}$$ does not appear to be used? Is $c_n$ derieved from that? How? 
UPDATE
In another example I have: 
$$\sum_{n=1}^{\infty} \frac{n^2(x-3)^{n+1}}{5^n}, \qquad c_{n+1}=\frac{n^2}{5^n}$$
If I try to express the series in 
$$\sum_{n=1}^{\infty} c_n (x-a)^n$$ 
I get 
$$\frac{n^2 (x-3)}{5^n} \cdot (x-3)^n, \qquad c_n = \frac{n^2 (x-3)}{5^n}$$
Or can the $(x-3)$
 be removed somehow?
 [/INST] Hint: 
Let $$a_{0}+a_{1}z+a_{2}z^{2}+a_{3}z^{3}+a_{4}z^{4}+\cdots$$ be a power-series, and consider the series  $$a_{1}+a_{2}z^{1}+a_{3}z^{2}+a_{4}z^{3}+\cdots$$ which is obtained by differentiating the power series term by term. The derived series has the same circle of convergence as the original series. Similarly the series $$\sum _{n =0}^{\infty }\dfrac {a_{n}z^{n+1}} {n+1},$$ obtained by integrating the original power-series term by term, has the same circle of convergence as $\sum _{n=0}^{n=\infty }a_{n}z^{n}$.
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