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?

This is true.
1) Pick $\epsilon>0$. Then there exists $N$ such that
$$
\sqrt[n]{a_n}\leq 1+\epsilon\quad\Rightarrow\quad a_n\leq (1+\epsilon)^n\qquad\forall n\geq N.
$$
Now 
$$
\sum_{k=1}^Ka_k =\sum_{k=1}^{N-1}a_k+\sum_{k=N}^Ka_k\leq C+\sum_{k=N}^K(1+\epsilon)^k=C+ (K-N+1)(1+\epsilon)^K\leq C+K(1+\epsilon)^K
$$
where $C=\sum_{k=1}^{N-1}a_k$ is fixed.
So
$$
\sqrt[K]{\sum_{k=1}^Ka_k }\leq (1+\epsilon)\left(C +\frac{K}{(1+\epsilon)^K}\right)^\frac{1}{K} \longrightarrow 1+\epsilon.
$$
This proves that 
$$
\limsup \sqrt[K]{\sum_{k=1}^Ka_k }\leq 1+\epsilon\quad\forall\epsilon>0\quad\Rightarrow \quad\limsup \sqrt[K]{\sum_{k=1}^Ka_k }\leq 1.
$$
2) Take a subsequence such that 
$$
\lim \sqrt[n_k]{a_{n_k}}=1.
$$
Pick $\epsilon>0$. There exists $K$ such that
$$
\sqrt[n_k]{a_{n_k}}\geq 1-\epsilon\quad\Rightarrow \quad a_{n_k}\geq (1-\epsilon)^{n_k}\quad\forall k\geq K.
$$
Now 
$$
\sum_{n=1}^{n_k}a_n\geq a_{n_K}\geq (1-\epsilon)^{n_k}\quad\forall k\geq K.
$$
So
$$
\sqrt[n_K]{\sum_{n=1}^{n_k}a_n}\geq 1-\epsilon \quad\forall k\geq K.
$$
Hence
$$
\limsup \sqrt[N]{\sum_{n=1}^{N}a_n}\geq 1-\epsilon\quad\forall\epsilon>0\quad \limsup \sqrt[N]{\sum_{n=1}^{N}a_n}\geq 1.
$$
Both inequalities are  now proven, so
$$
\limsup \sqrt[N]{\sum_{n=1}^{N}a_n}= 1.
$$
