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I'm stuck trying to understand a proof of the following theorem: Let $ \sum a_nx^n$ be a power series with radius of convergence $ R $. Then $ \sum na_nx^{n-1}$ also has radius of convergence $ R $.

$ \sum_{n=1}^{\infty}na_nx^{n-1} = \sum_{n=0}^{\infty}(n+1)a_{n+1}x^n$. Then the proof in my lecture notes states that $ \limsup \sqrt[n]{(n+1)|a_{n+1}|} = \limsup \sqrt[n]{|a_n|} = \frac{1}{R} $.

I fail to understand where that equality between upper limits comes from. Obviously, $ \limsup \sqrt[n]{n} = 1$ and we could break down the limit into a product, but a product of upper limits is $ \le $ than the upper limit of a product, so I still can't see how we would obtain an equality.

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    $\begingroup$ In this case, the upper limit of $n^{1/n}$ is a plain non-zero limit. $\endgroup$ – André Nicolas Jun 9 '14 at 16:39
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You are correct that

$$\limsup(n|a_n|)^{1/n}\leq \limsup(|a_n|)^{1/n}\limsup(n)^{1/n}=\limsup(|a_n|)^{1/n}.$$

But for all $n$

$$|a_n| \leq n|a_n|$$

so

$$\limsup(|a_n|)^{1/n}\leq \limsup(n|a_n|)^{1/n}$$

and we have equality.

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Refer to the wikipedia section on root test and ROC.

Prove that the radius of convergence is always $R = \dfrac{1}{\limsup \sqrt[n]{|c_n|}}$. Then your equality.

If there were a larger radius of convergence of $\sum c_n (z - a)^n$, say $R' \gt R$, then it is not true of $R$ that the series converges for all $|z - a| \lt R$ and diverges for all $|z - a| \gt R$ since now we've assumed the same for $R'$. Contradiction.

Thus, in particular ROC $R$ is unique, but for your problem it always equals that formula $\square$

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