Prove that the sequence $a_{1}= 1$, $a_{n+1} = \sqrt[n]{a_{1}+\dots+a_{n}}$ is bounded below So I have a sequence $a_{n+1} = \sqrt[n]{a_{1}+\dots+a_{n}}$, $n \in \mathbb{N}$ and where is $a_{1}= 1$.
I have to prove that there is $c > 0$ such that for every $n \in \mathbb{N}$,  $a_{n} \geq c$ holds.
I began proving this with mathematical induction and I came to this point for $n \rightarrow n+1$ $$a_{1}\dots+a_{n} \geq c^n$$ and I do not know how to finish it.
Also I have to prove that this is increasing sequence and I came to this point $$a_{1}+\dots+a_{n}+a_{n+1}>=a_{1}+\dots+a_{n}$$ and from this I get that $$a_{n+1}>=0$$
If is this even correct so far?
Any help?
 A: By induction,
$$a_1\ge1$$ and $$\forall k\in[1,n]:a_k\ge1\implies a_1+a_2+\cdots a_n\ge1\implies  a_{n+1}\ge1.$$
This bound is tight.
A: So, we have $a_1=1,a_2=\sqrt{a_1}=1,a_3=\sqrt[2]{a_1+a_2}=\sqrt{2}$,we guess that $a_n\geq 1,\forall n\in\mathbb{N}$.
We'll prove our proposition using Induction


*

*For $n=1$, it's clear that $a_1=1\geq 1$

*We'll assume that for $1\leq k\leq n, a_n\geq 1$,then we get $$a_n=\sqrt[n-1]{a_1+a_2+\cdots+a_{n-1}}\geq 1 (1)$$ and $$a_1+a_2+\cdots +a_{n-1}\geq 1 (2)$$

*For $n=k+1$, we get $$a_{n+1}=\sqrt[n]{a_1+a_2+a_3+\cdots a_{n-1}+a_n}=\sqrt[n]{a_1+a_2+a_3+\cdots a_{n-1}+\sqrt[n-1]{a_1+a_2+\cdots+a_{n-1}}}\geq \sqrt[n]{1+1}\geq \sqrt[n]{1}=1$$
So, we have $a_n\geq 1$, for all $n\in\mathbb{N}$, hence $a_n$ is bounded below by $1$.


Furthermore, $1$ is even the greatest lower bound a.k.a infimum ${a_n}$.
Take any arbitrary $\epsilon>0$, there exists $n=1$ such that $$a_n-1=1-1=0<\epsilon$$ hence, inf $(a_n)_{n\in\mathbb{N}}=1$.
Hope this will help you!
A: In this posting, we will show that in fact $a_n\xrightarrow{n\rightarrow\infty}1$ converges for any initial condition $a_1=x>0$.
For $x>0$, define $a_1(x)=x$ and
$$a_{n+1}(x)=\Big(a_1(x)+\ldots + a_n(x)\Big)^{1/n}$$
Using the fact that $x^{1/m}\leq \max(1,x)$ for all $m\in\mathbb{N}$, we obtain
$$\begin{align}
(a_1(x))^{1/(n-1)}\leq a_n(x)\leq a_n(1)(\max(1,x)^{1/(n-1)})\tag{1}\label{one}
\end{align}$$
for $n\geq 2$. Indeed, the claim holds or $n=2$. Assuming the statement holds for $a_j(x)$, $2\leq j\leq n$ yields
$$\begin{align}
a_{n+1}(x)&\leq \big(a_1(1)x +a_2(1)x+\ldots +a_n(1)\max(1,x^{1/(n-1)})\big)^{1/n}\\
&\leq\Big(\big(a_1(1)+\ldots a_n(1)\big)\max(1,x)\Big)^{1/n}\\
&=a_{n+1}(1)\max(1,x^{1/n})
\end{align}
$$

Consider the case $x=1$. Clearly $a_n(1):=a_n\geq 1$ for all $n\in\mathbb{N}$. By induction
$$a_n\leq 2,\qquad n\in\mathbb{N}$$
Indeed, for $n=1$ this holds trivially. Assuming that $a_k\leq 2$ for $1\leq n$, gives
$$a_{n+1}=(a_1+\ldots+a_n)^{1/n}\leq (2n)^{1/n}\leq (2^n)^{1/n}=2$$
Consequently,
$$1\leq a_{n+1}\leq2^{1/n}2^{1/n}\xrightarrow{n\rightarrow\infty}1$$
Notice that if $a_n(1)$ converges and $a_1=1$,  then from \eqref{one}
$$\lim_na_n(x)=\lim_na_n(1)=1$$
