Radius of convergence for "shifted sequences" in proof of derivative of power series The Hadamard formula for the radius of convergence of a power series $\sum_j c_j z^j$ gives that
$$1/R = \limsup_j |c_j|^{1/j}.$$
The power series giving the derivative of a function is
$\sum_{j=1}^\infty jc_jz^{j-1}$. The proof of this statement, in e.g. Ahlfors, relies on saying that the radius of convergence of this series is the same as the radius of convergence of the original series, and the argument as given show that
$$\limsup_j |jc_j|^{1/j} = \limsup_j |c_j|^{1/j}.$$
But the actual radius of convergence of the "derived series" is not the LHS of the above, but rather
$$\limsup_j |(j+1)c_{j+1}|^{1/j}.$$
It is easy enough to see that this is equal to
$$\limsup_j |c_{j+1}|^{1/j},$$
but why is this equal to
$$\limsup_j |c_{j}|^{1/j}?$$
 A: We first show some lemmas.
Lemma 1:
If $a_n \geq 0$, $a_n \to a$, and $b_n \to 1$ where $a_n, b_n$ are in $\mathbb R$, then $a_n^{b_n} \to a$.
Proof: $b_n$ is eventually nonnegative, so we can just suppose $b_n$ is nonnegative. We have that for each $\epsilon >0$, there is some $N$ s.t. for all $n \geq N$, $a - \epsilon < a_n < a+\epsilon$. Then $x \mapsto x^{b_n}$ is order preserving for all $n$ since $b_n \geq 0$. Thus $(a-\epsilon)^b_n < a_n^{b_n} < (a+\epsilon)^{b_n}$ for all $n \geq N$.
Then note that $(a+\epsilon)^{b_n} \to a+\epsilon$ as $x\mapsto (a+\epsilon)^x$ is continuous. Thus
$$\limsup_n a_n^{b_n} \leq \limsup_n (a+\epsilon)^{b_n} = a+\epsilon,$$
for all $\epsilon > 0$. Thus $\limsup_n a_n^{b_n} \leq a$. Note also that $0 \leq a_n^{b_n}$ so if $a=0$, then $a_n^{b_n} \to 0 = a$. Thus we are done in the case $a=0$.
Now suppose $a \ne 0$. Then we can take $\epsilon$ small enough s.t. $0 < a-\epsilon$. Then we also have $(a-\epsilon)^{b_n} \to a-\epsilon$. Then
$$\liminf_n a_n^{b_n} \geq \liminf_n (a-\epsilon)^{b_n} = a-\epsilon,$$
for any $\epsilon$ sufficiently small. Thus $\liminf_n a_n^{b_n} \geq a$. Putting this together with the result above finishes the proof.
Lemma 2:
If $a_n \geq 0$ and $b_n \to 1$ where $a_n, b_n$ are in $\mathbb R$, then $\limsup_n a_n^{b_n} = \limsup_n a_n$.
Proof: Since $b_n$ eventually $> 0$, we can WLOG suppose $b_n \in [0,\infty)$. Recall that the limsup is just the sup of the limit points (working in the extended reals, so we can have $\infty$ as a limit point). Thus it suffices to show that the sets of limit points of $a_n^{b_n}$ and $a_n$ are the same. Call the $S_b, S$ resp.
Let $L \in S_b$. Then there is a subsequence $a_{n_j}^{b_{n_j}} \to L$. Either $L < \infty$ or $L =\infty$. First suppose $L < \infty$. Then note that $1/b_{n_j} \to 1$, so by the lemma above,
$$(a_{n_j}^{b_{n_j}})^{1/b_{n_j}} = a_{n_j} \to L$$
so $L \in S$. Now suppose $L = \infty$. Then for $j$ large enough, $b_{n_j} < 2$. Thus
$$a_{n_j} = (a_{n_j}^{b_{n_j}})^{1/b_{n_j}} \geq \sqrt{a_{n_j}^{b_{n_j}}} \to \infty.$$
so $L \in S$ also. Thus $S_b \subseteq S$.
The reverse inclusion is very similar. Thus $S = S_b$ and the result follows.
Main result:
We now can easily show the main result.
$$\limsup_j |c_j|^{1/j} = \limsup_j |c_{j+1}|^{1/(j+1)} = \limsup_j (|c_{j+1}|^{1/j})^{j/(j+1)} = \limsup_j |c_{j+1}|^{1/j},$$
where all but the last equality are trivial and the final equality follows from lemma 2.
