Does $\limsup_{n\to\infty}\sqrt[n]{a_n}=1$ imply $\limsup_{n\to\infty}\sqrt[n]{a_1+\cdots+a_n}=1$ for positive real sequences [duplicate]

I am working on an exercise from calculus:

Let $\{a_n\}_{n\ge 1}$ be a positive sequence such that $\limsup_{n\to\infty}\sqrt[n]{a_n}=1$.
Show that $\limsup_{n\to\infty}\sqrt[n]{a_1+\cdots+a_n}=1$.

Here is my attempt:

Trivially we have$$\mathop {\lim\mathrm{sup}} \limits_{n\rightarrow \infty}\sqrt[n]{a_1+\cdots +a_n}\ge \mathop {\lim\mathrm{sup}} \limits_{n\rightarrow \infty}\sqrt[n]{a_n}=1.~~~~~~~~~(1)$$

For the converse inequality, by definition of $\limsup$, for any $\epsilon>0$ there exists some $N>0$ such that $\sqrt[n]{a_n}<1+\epsilon$, hence$$\begin{aligned}\sqrt[n]{\frac{a_1+\cdots +a_n}{n}}&\le \sqrt[n]{\frac{a_1+\cdots +a_N}{n}}+\sqrt[n]{\frac{a_{N+1}+\cdots +a_n}{n}}~~~~~~~~~(2)\\&\le \sqrt[n]{\frac{a_1+\cdots +a_N}{n}}+\sqrt[n]{\frac{\left( n-N \right) \left( 1+\epsilon \right) ^n}{n}}\\&\le \sqrt[n]{\frac{a_1+\cdots +a_N}{n}}+\left( 1+\epsilon \right) \cdot \sqrt[n]{\frac{n-N}{n}},\end{aligned}$$however$$\mathop {\lim\mathrm{sup}} \limits_{n\rightarrow \infty}\sqrt[n]{a_1+\cdots +a_n}=\mathop {\lim\mathrm{sup}} \limits_{n\rightarrow \infty}\sqrt[n]{\frac{a_1+\cdots +a_n}{n}}\le 2+\epsilon ,~~~~~~~~~(3)$$which is larger than expected.