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


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


you only need


and this is why that book does what they did:


and now you only need to convince yourself (meaning: prove it) that

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

  • $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ 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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