Let $\sum c_k x^k$ be a power series with radius of convergence $R$. Then the integral series $$\sum_{k=0}^\infty \frac{c_k}{k + 1}x^{k+1}$$ also has radius of convergence $R$.
I'm reading Real Mathematical Analysis by Pugh and here's how he justifies this statement (chapter 4, theorem 12):
The radius of convergence of the integral series is determined by the exponential growth rate of its coefficients, $$\limsup_{k\to\infty} \sqrt[k]{\left|\frac{c_{k-1}}{k}\right|} = \limsup_{k\to\infty}(|c_{k-1}|^{1/(k-1)})^{(k-1)/k}\left(\frac{1}{k}\right)^{1/k}.$$ Since $(k − 1)/k \to 1$ and $k^{−1/k} \to 1$ as $k \to \infty$, we see that the integral series has the same radius of convergence $R$ as the original series.
The problem is, if $(k-1)/k \to 1$ how do we know that $$\limsup_{k\to\infty}(|c_k|^{1/k})^{(k-1)/k} = \limsup_{k\to\infty}|c_k|^{1/k}?$$ In general $\limsup a_n = a$, $\lim b_n = b$ does not imply $\limsup a_n^{b_n} = a^b$. Even worse: $b_n \to 1$ does not imply $\limsup a_n^{b_n} = a$. For example, take $b_n = 2n/(2n + 2)$, $a_n = -1$. Then $a^b = -1$, but $\limsup a_n^{b_n} = 1$.
I feel that my example is somewhat pathological and under certain assumptions $\limsup a_n = a$, $\lim b_n = b$ should imply $\limsup a_n^{b_n} = a^b$. So, what are these assumptions and how to prove this statement?
You need to evaluate
$$\lim_{k\to\infty}\sup\sqrt[k]{\left|\frac{c_k}{k+1}\right|}=\lim_{k\to\infty}\sup\sqrt[k]{\left|\frac{c_k}{k+1}\right|}=\lim_{k\to\infty}\sup\sqrt[k]{|c_k|}\cdot\lim_{k\to\infty}\sqrt[k]{\frac1{k+1}}$$
The last equality is justified since both lim sup exist, though the right one is just limit as it exists and equals one. This is the end and I can't understand why he had to do such messing calculations since it is the same aking what I did or any $\;\frac{c_m}{m+1}\;$ ...
Added Since
$$\lim_{n\to\infty}\sqrt[k]k=\lim_{n\to\infty}\sqrt[k]{k+1}=1$$
you only need
$$\sqrt[k]{|c_{k-1}|}=\sqrt[k]{|c_k|}$$
and this is why that book does what they did:
$$\sqrt[k]{|c_{k-1}|}=\left(\sqrt[k-1]{|c_{k-1}|}\right)^{\frac{k-1}k}$$
and now you only need to convince yourself (meaning: prove it) that
$$a_n\xrightarrow[n\to\infty]{}1\implies a_n^{x_n}\xrightarrow[n\to\infty]{}1\;\;\text{ whenever}\;\;x_n\to1$$