Prove that a power series and its deriviative has the same convergence radius Let $ \sum_{n=0}^{\infty}a_{n}x^{n} $ be a power series. prove that
$ \sum_{n=0}^{\infty}a_{n}x^{n},\sum_{n=0}^{\infty}\left(n+1\right)a_{n+1}x^{n} $
has the same convergence radius.
So, let $ \limsup\sqrt[n]{|a_{n}|}=\beta $
It follows that $ \limsup\sqrt[n+1]{|a_{n+1}|}=\beta $ (Im not sure about this argument, we'll sign it as (*) )
Its enough to show that $ \limsup\sqrt[n]{\left(n+1\right)|a_{n+1}|}=\beta $
Let $ \limsup\sqrt[n]{\left(n+1\right)|a_{n+1}|}=\gamma $ and we'll show that $\beta=\gamma $
From the assumption above, we can find a subsequence such that $ \lim_{k\to\infty}\sqrt[n_{k}]{|a_{n_{k}}|}=\beta$
And we can notice that :
$ \lim_{k\to\infty}\sqrt[n_{k}]{\left(n_{k}+1\right)|a_{n_{k}+1}|}=\lim_{k\to\infty}\left(n_{k}+1\right)^{\frac{1}{n_{k}}}\cdot\lim_{k\to\infty}\left(|a_{n_{k}+1}|^{\frac{1}{n_{k}+1}}\right)^{\frac{n_{k}+1}{n_{k}}} $
and because of (*)
$ \lim_{k\to\infty}\left(n_{k}+1\right)^{\frac{1}{n_{k}}}\cdot\lim_{k\to\infty}\left(|a_{n_{k}+1}|^{\frac{1}{n_{k}+1}}\right)^{\frac{n_{k}+1}{n_{k}}}=\beta $
So $ \beta \leq \gamma $
Now, we'll take a subsequence such that $ \lim_{j\to\infty}\sqrt[n_{j}]{\left(n_{j}+1\right)|a_{n_{j}+1}}=\gamma $
And we'll notice that
$ \gamma=\lim_{j\to\infty}\sqrt[n_{j}]{\left(n_{j}+1\right)|a_{n_{j}+1}|}=\lim_{j\to\infty}\left(n_{j}+1\right)^{\frac{1}{n_{j}}}\cdot\lim_{j\to\infty}|a_{n_{j}+1}|^{\frac{1}{n_{j}}}=\lim_{j\to\infty}\left(|a_{n_{j}+1}|^{\frac{1}{n_{j}+1}}\right)^{\frac{n_{j+1}}{n_{j}}}=\gamma $
Therefore, because of (*)
$ \lim_{j\to\infty}|a_{n_{j}+1}|^{\frac{1}{n_{j}+1}}=\lim_{j\to\infty}|a_{n_{j}}|^{\frac{1}{n_{j}}}=\gamma $
And thus $ \gamma \leq \beta $ and we get that $ \gamma = \beta $
Now everything depend's on the validity of (*) So I hope you could tell me if it's okay and maybe help my justify it.
I feel like it's abuse of notation because $ a_{n_{k+1}}\neq a_{n_{k}+1} $ so  I dont know if my argument holds.
Thanks in advance
 A: You don't say where your indices start, so I pick $1$.  Altering this is easy.
$$  \left( \sqrt[n]{|a_n|} \right)_{n \geq 1} = (|a_1|, \sqrt{a_2}, \sqrt[3]{a_3}, \dots )  \text{ and }  $$
$$  \left( \sqrt[n+1]{|a_{n+1}|} \right)_{n \geq 1} = (\sqrt{a_2}, \sqrt[3]{a_3}, \dots )  \text{,}  $$
so the second sequence is the subsequence of the first sequence formed by deleting the first term.  The limit you ask about is indifferent to any finite segment of the sequence not containing a maximum of the sequence, so the two limits agree as long as the first term is not a maximal term of the sequence.
A: Based on this general result (the equality part), we indeed have:
$$ 
\limsup_{n\to\infty}\sqrt[n]{(n+1)|a_{n+1}|}=\lim_{n\to\infty}\sqrt[n]{n+1}\limsup_{n\to\infty}\sqrt[n]{|a_{n+1}|} $$
You are basically attempting to prove it by choosing special (their limits are hitting the $\limsup$'s) subsequences on both sides, but in the process you set $\limsup$ of a sequence to $\limsup$ of one of its non-special sequences (which is wrong as stated in the comments on the other answer). The limit on non-special sequences might not even exist. I think you do this by assuming that the special subsequence for the product of sequences are also special for the individual sequences (and vice versa).
Slightly cleaner application of the general result:
$$ \sum_{n\geq 0} (n+1)a_{n+1} x^n = \sum_{n\geq 1} na_n x^{n-1}  $$
The convergence radius of $ \sum_{n\geq 1} na_n x^{n-1} $ is the same as the one of
$$ \sum_{n\geq 1} na_n x^{n} = x \sum_{n\geq 1} na_n x^{n-1}. $$
Then:
$$ 
\limsup_{n\to\infty}\sqrt[n]{n|a_{n}|}=\lim_{n\to\infty}\sqrt[n]{n}\limsup_{n\to\infty}\sqrt[n]{|a_n|} = \limsup_{n\to\infty}\sqrt[n]{|a_n|}.$$
