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?
Since $\{a_n\}$ is a positive sequence, we have $$ \limsup_{n\to\infty}\sqrt[n]{a_1+a_2+\ldots+a_n}\ge\limsup_{n\to\infty}\sqrt[n]{a_n}=1 $$ For all $\epsilon>0$, there is a positive integer $N$, so that for all $n\gt N$, $$ a_n\le(1+\epsilon)^n $$ Thus, $$ \begin{align} &\limsup_{n\to\infty}\sqrt[n]{a_1+a_2+\ldots+a_n}\\ &=\limsup_{n\to\infty}\left(\sum_{k=1}^Na_k+\sum_{k=N+1}^na_k\right)^{1/n}\\ &\le\limsup_{n\to\infty}\left(\sum_{k=1}^Na_k+(n-N)(1+\epsilon)^n\right)^{1/n}\\ &=\limsup_{n\to\infty}\left(1+\frac{\sum_{k=1}^Na_k}{(n-N)(1+\epsilon)^n}\right)^{1/n}(n-N)^{1/n}(1+\epsilon)\\[6pt] &=1\cdot1\cdot(1+\epsilon)\\[16pt] &=1+\epsilon \end{align} $$ Therefore, since $\epsilon$ was arbitrary, $$ \limsup_{n\to\infty}\sqrt[n]{a_1+a_2+\ldots+a_n}=1 $$
In general, let
$$\alpha = \limsup_{n\to\infty} \sqrt[n]{a_n}$$
for the sequence $\{ a_n \}$ of non-negative real numbers such that not every term is equal to zero. Then we have
$$ \limsup_{n\to\infty} \sqrt[n]{a_1 + \cdots + a_n} = \max \{ \alpha, 1 \}. $$
Proof using elementary analysis.
Case 1. Assume $\alpha \geq 1$. On the one hand, we have
$$ \limsup_{n\to\infty} \left( \sum_{k=1}^{n} a_k \right)^{1/n} \geq \limsup_{n\to\infty} \sqrt[n]{a_n} = \alpha. $$
Thus if $\alpha = \infty$, then this automatically implies the assertion. This shows that we assume without losing the generality that $\alpha < \infty$.
For any $\epsilon > 0$, there exists $N$ such that $a_n \leq (\alpha + \epsilon)^{n}$ for all $n \geq N$. Since
$$ \sum_{k=1}^{n} a_k \leq \left( \sum_{k=1}^{N} a_k \right) + (\alpha + \epsilon)^{N+1} \cdot \frac{(\alpha + \epsilon)^{n-N} - 1}{(\alpha + \epsilon) - 1} $$
and $\alpha + \epsilon > 1$, it follows that
$$ \limsup_{n\to\infty} \left( \sum_{k=1}^{n} a_k \right)^{1/n} \leq \alpha + \epsilon. $$
As this is true for any $\epsilon > 0$, we have
$$ \limsup_{n\to\infty} \left( \sum_{k=1}^{n} a_k \right)^{1/n} \leq \alpha. $$
This complete s the proof for the case $\alpha \geq 1$.
Case 2. Assume $\alpha < 1$. Then for some $\alpha < \beta < 1$, there exists $C > 0$ such that $a_n \leq C \beta^{n}$ for all $n$. This shows that
$$ \sum_{k=1}^{n} a_k \leq \sum_{k=1}^{n} C \beta^{k} \leq \frac{C\beta}{1-\beta} < \infty. $$
On the other hand, since $a_n > 0$ for some $n$, it follows that
$$ 0 < C' \leq \sum_{k=1}^{n} a_k $$
for some $C' > 0$ for sufficiently large $n$. This shows that
$$ \limsup_{n\to\infty} \sqrt[n]{a_1 + \cdots + a_n} = 1. $$
Proof using complex analysis.
Let
$$ f(z) = \sum_{n=0}^{\infty} a_n z^n $$
and $R$ be the radius of convergence of this series. Then we have
$$ \frac{1}{R} = \limsup_{n\to\infty} \sqrt[n]{a_n} = \alpha,$$
where we adopt the convention that $1/\infty = 0$ and $1/0 = \infty$. But since
$$ \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} a_k \right) z^{n} = \frac{f(z)}{1-z} $$
has the radius of convergence as $\min \{R, 1 \}$ because of the singularity at $z = 1$, it follows that
$$ \limsup_{n\to\infty} \sqrt[n]{a_1 + \cdots + a_n} = \frac{1}{\min \{1, R \}} = \max \{1, \alpha \}.$$
$\sum_1^{n} a_k\geq a_n$
So,$\limsup\sqrt[n]{\sum_1^{n} a_k}\geq1$
Choose N such that $n>N$, $ a_n\leq(1+\epsilon)^n$.
Let $\sum_{k=1}^Na_k=A$
$$\sum_{k=1}^na_k\leq A+(1+\epsilon)^N\dfrac{(1+\epsilon)^{n-N+1}-1}{\epsilon}$$ $$\sqrt[n]{\sum_{k=1}^na_k}\leq(1+\epsilon)( (A -\frac 1\epsilon)(1+\epsilon)^{-n}+1)^{\frac 1 n}$$ Taking the limit we get,$\limsup\sqrt[n]{\sum_{k=1}^na_k}\leq1+\epsilon$, Proving the result.
\varlimsup instead of \limsup might have confused some people. I know that I missed the bar above the $\lim$ when I read the question.
– robjohn
Mar 07 '13 at 13:55
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 $$ \limsup \sqrt[N]{\sum_{n=1}^{N}a_n}= 1. $$
Let $\epsilon>0$. Then $ a_n<(1+\epsilon)^n, \forall n\geq n_0 $ for some $n_0\in\mathbb N$. Therefore $$1\leq\limsup_{n\rightarrow\infty}\sqrt[n]{a_{1}+a_{2}+\cdots+a_{n}}\leq 1+\epsilon.$$
