If $\limsup_n\sqrt[n]{a_n}=\frac{1}{r}$, then $\limsup_n\sqrt[n]{(n+1)a_{n+1}}=?$ 
If $\limsup_n\sqrt[n]{|a_n|}=\frac{1}{r}$, then $\limsup_n\sqrt[n]{|(n+1)a_{n+1}|}=?$

Can I separate the product of the second limit as $$\limsup_n\sqrt[n]{|(n+1)a_{n+1}|}=\limsup_n\sqrt[n]{(n+1)}\limsup_n\sqrt[n]{|a_{n+1}|}$$since the first limit exists and $\sqrt[n]{(n+1)}$ is bounded, if not does the RHS majorize LHS, so that I deduce the limit is also $\frac{1}{r}$ 
(Original Exercise: If $\sum_na_nz^n$ has radius of convergence $r$, so has $\sum_n na_nz^{n-1}$)

It is not true, in general, that if $x_n,y_n$ are sequences, then $$\limsup x_ny_n = (\limsup x_n)(\limsup y_n).$$
For example, if $$x_n=\begin{cases}1&n\text{ even}\\2&n\text{ odd}\end{cases}$$ and $y_n=\frac{2}{x_n}$, then $\limsup x_n=\limsup y_n = 2$, but $\limsup x_ny_n = 2\neq 2\cdot 2$.
But if $x_n$ converges to a positive value, it is true.
