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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?

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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

$$a_n\xrightarrow[n\to\infty]{}1\implies a_n^{x_n}\xrightarrow[n\to\infty]{}1\;\;\text{ whenever}\;\;x_n\to1$$

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  • $\begingroup$ We have to do these calculations because $k$th coefficient of the integral series is $c_{k-1}/k$, not $c_k/(k+1)$. We don't know that $\limsup \sqrt[k]{|c_{k-1}/k|} = \limsup \sqrt[k]{|c_k/(k + 1)|}$. We can't just shift $k$ by 1 since two sequences are different. $\endgroup$ – edubrovskiy Jul 14 '16 at 12:04
  • $\begingroup$ Well, maybe $\limsup \sqrt[k]{|c_{k-1}/k|} = \limsup \sqrt[k]{|c_k/(k + 1)|}$ but I don't know how to prove it. $\endgroup$ – edubrovskiy Jul 14 '16 at 12:13
  • $\begingroup$ If you read the last part of my comment it really doesn't matter, but I can understand that book shows the most formal, accurate way a,dn that's why they do that...and yes: we can justify the shifting as it really you can always multiply the whole series by $\;x\;$ or divide by $\;x\neq0\;$ , and that way the exponent gets shifted. $\endgroup$ – DonAntonio Jul 14 '16 at 12:38

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