Is $\limsup \left(|a_n|^{1/n}\right)=\limsup \left(|a_n|^{1/(n-1)}\right)$? How does one go about proving that :
$$\limsup\left(|a_n|^\frac{1}{n}\right)=\limsup\left(|a_n|^\frac{1}{n-1}\right)$$
without the use of logarithms? 
 A: Note that $a_n^{\frac{1}{n-1}}=(a_n^{\frac{1}{n}})^{\frac{n}{n-1}}=(a_n^{\frac{1}{n}})^{\frac{1}{1-\frac{1}{n}}}$
What happens when $n\to \infty$?
A: Since $\limsup$ can be realized to the subsequence, it amounts to deal with the question that $\lim a_{n}^{1/n}=\lim a_{n}^{1/(n-1)}$ for nonnegative $a_{n}$.
A quick note: $\limsup a_{n}^{1/n}=\lim_{k}a_{n_{k}}^{1/n_{k}}=L$. Either $L<\infty$ or $L=\infty$. In the following, without loss of generality, we stick to the notation $n$ instead of $n_{k}$ and we deal with $L<\infty$. For $L=\infty$ can be treated similarly. 
Now let $L=\lim a_{n}^{1/n}$. Assume $L>0$ first. Let $\epsilon>0$ be so small that $L-\epsilon>0$. Then for sufficiently large $n$ we have $L-\epsilon<a_{n}^{1/n}<L+\epsilon$, so $(L-\epsilon)^{n/(n-1)}<a_{n}^{1/(n-1)}<(L+\epsilon)^{n/(n-1)}$.
We are to show that $c^{n/(n-1)}=c\cdot c^{1/n-1}\rightarrow c$ for $c>0$. Or we just need to show that $c^{1/(n-1)}\rightarrow 1$. Assume that $c>1$, write $c=1+p$, as $(1+p/(n-1))^{(n-1)}\geq 1+(n-1)(p/(n-1))=1+p$, then $(1+p)^{1/(n-1)}\leq 1+p/(n-1)$. Now $(1+p)^{1/(n-1)}\geq 1$, taking $n\rightarrow\infty$ and use Squeeze Theorem gets the result. For $0<c<1$, one considers $1/c$.
Back to $(L-\epsilon)^{n/(n-1)}<a_{n}^{1/(n-1)}<(L+\epsilon)^{n/(n-1)}$, use Squeeze Theorem again to get $L-\epsilon\leq\liminf a_{n}^{1/(n-1)}\leq\limsup a_{n}^{1/(n-1)}\leq L+\epsilon$. Now let $\epsilon\downarrow 0$ and so $\lim a_{n}^{1/(n-1)}=\limsup a_{n}^{1/(n-1)}=\liminf a_{n}^{1/(n-1)}=L$.
For $L=0$, use $\epsilon$-argument and the similar technique above.
Thanks to @user334639 pointing out the issue of realization to subsequences: First, realize $\limsup a_{n}^{1/n}=\lim_{k}a_{n_{k}}^{1/n_{k}}=\lim_{k}a_{n_{k}}^{1/(n_{k}-1)}\leq\limsup a_{n}^{1/(n-1)}$. Then, realize $\limsup a_{n}^{1/(n-1)}=\lim_{l}a_{l}^{1/(n_{l}-1)}=\lim_{l}a_{l}^{1/n_{l}}\leq\limsup a_{n}^{1/n}$. 
