Show $ \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{1}+a_{2}+\cdots+a_{n}}}=1 $ if $\displaystyle \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{n}}}=1 $ Let $\{a_{n}\}$ be a positive sequence with $\displaystyle \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{n}}}=1 $. How can we show that
$$ \varlimsup_{n\rightarrow\infty}{\sqrt[n]{a_{1}+a_{2}+\cdots+a_{n}}}=1 $$
I am not sure the problem is true. If it is false, what is the counterexample?

Let $\epsilon>0$. Then $ a_n<(1+\epsilon)^n, \forall n\geq  n_0 $  for some $n_0\in\mathbb N$. Therefore
$$1\leq\limsup_{n\rightarrow\infty}\sqrt[n]{a_{1}+a_{2}+\cdots+a_{n}}\leq 1+\epsilon.$$
