Question about two sequences with a common limit Suppose $a _n$ is a sequence of positive integers such that $ \lim_{n \rightarrow \infty} (a_n)^{\frac{1}{n}} $ exists.
Suppose there exists a sequence of positive integers $ b_n $ such that
$$
a_n = \sum_{i=0}^{n-1} b_i
$$
Is there a good way to see why $ \lim_{n \to \infty} (a_n)^{\frac{1}{n}} = \lim_{n \to \infty} (b_n)^{\frac{1}{n}} $?
 A: As $\{a_n\}$ is a sequence of positive integers, division by $a_n$ doesn't cause any problem. We use $\lim$ to denote $\lim_{n\to+\infty}$.

$$1<\lim {a_{n+1}\over a_n}<+\infty\implies\lim a_n^{1\over n}=\lim b_n^{1\over n}$$.

Proof: $$1<l:=\lim {a_{n+1}\over a_n}<+\infty\\
b_n=a_{n+1}-a_n=a_n\left({a_{n+1}\over a_n}-1\right)
\implies b_n^{1\over n}=a_n^{1\over n}\left({a_{n+1}\over a_n}-1\right)^{1\over n}\quad\cdots(1)\\
$$
Given $0<\varepsilon< l-1$, for large enough $n$ we have
$$
(l-\varepsilon)\le {a_{n+1}\over a_n}\le(l+\varepsilon)
\implies (l-\varepsilon-1)^{1\over n}\le\left({a_{n+1}\over a_n}-1\right)^{1\over n}\le(l+\varepsilon-1)^{1\over n}
$$
Applying squeeze theorem we see $\lim \left({a_{n+1}\over a_n}-1\right)^{1\over n}=1$. So $(1)$ suggests $\lim b_n^{1\over n}$ exists and is equal to $\lim a_n^{1\over n}$.

Counter-example

So in order to provide a counter-example we're going to pick $\{a_n\}$ in such a way that $\lim {a_{n+1}\over a_n}$ fails to exist. The following works
$$\begin{align}
&b_n=\begin{cases}
2,&n=0\\
1, &n\text{ is even and positive}\\
{2^{n+1}}-{2^{n-1}}, &n\text{ is odd}
\end{cases}\\
&a_n=\begin{cases}
2^n+{n\over 2}, &n\text{ is even}\\
2^{n-1}+{n+1\over2},&n\text{ is odd}
\end{cases}\end{align}
$$
In this case $\lim {a_{n+1}\over a_n}$ doesn't exist as
$$
\lim {a_{2n+1}\over a_{2n}}=1,\quad\lim {a_{2n+2}\over a_{2n+1}}=4
$$ 
$\lim a_n^{1\over n}=2$ and $\lim {b_n}^{1\over n}$ doesn't exist as 
$$\lim (2^{n+1}-2^{n-1})^{1\over n}=2\neq 1=\lim b_{2n}^{1\over 2n}$$ 
