Proving from a sequence of real numbers I am solving the following problem.
Let $a_n$ be a sequence of nonzero real numbers.
Assume that lim sup $|a_{n+1}/a_n|$ = L
a.  Let L’ be any number bigger than L. Prove that there exists N>0 such that $|a_{n+1}/a_n| < L’$ for any n > N.
b.  Prove that for any n>N we have |$a_n|<(L’)^{n-N}|a_N|$
For a. I know I want to use the definition of convergence which says for every epsilon greater than zero there exists a N>0 such that  n>N $\implies$ $|a_n-a|<\epsilon$ but my sequence is a little different and for that reason I am having a hard time with this problem. Thanks
 A: Because $\left|\frac{a_{n+1}}{a_n}\right| \to L$ and $L' > L$, we can find some $N \in \Bbb{N}$ such that
\begin{align*}
n > N &\implies \left|\left|\frac{a_{n+1}}{a_n}\right| - L\right| < L' - L \\
&\implies \left|\frac{a_{n+1}}{a_n}\right| - L < L' - L \\
&\implies \left|\frac{a_{n+1}}{a_n}\right| < L',
\end{align*}
proving (a). As a bit of intuition, I chose $\varepsilon = L' - L > 0$ because $L' - L$ is the distance between the limit $L$ and $L'$. So, when the sequence gets closer to $L$ than this distance, it has to lie below $L'$.
For part (b), we prove by induction on $n - N$. When $n - N = 1$, then
$$N + 1 > N \implies \implies \left|\frac{a_{N+1}}{a_N}\right| < L' \implies |a_{N+1}| < (L')^1 |a_N|.$$
So, let us suppose the given statement holds for $n - N = k$ for some $k \ge 1$. That is, for this $k$,
$$|a_{N+k}| < (L')^k |a_N|.$$
Then,
\begin{align*}
N + k + 1 > N &\implies \left|\frac{a_{N+k+1}}{a_{N+k}}\right| < L' \\
&\implies |a_{N+k+1}| < L'|a_{N+k}| \\
&\implies |a_{N+k+1}| < L'((L')^k |a_N|) \\
&\implies |a_{N+k+1}| < (L')^{k+1} |a_N|,
\end{align*}
completing the proof by induction.
A: First of all, the meaning of $\lim\sup \frac{a_{n+1}}{a_n}$:
Let us define a sequence $b_n=\sup_{k\geq n}\frac{a_{k+1}}{a_k}$. Then we'll find that:
$$\begin{align}b_1&=\sup\left\{\frac{a_2}{a_1},\frac{a_3}{a_2},...\right\}\\
b_2&=\sup\left\{\frac{a_3}{a_2},\frac{a_4}{a_3},...\right\}\\
.\\
.\\
b_k&=\sup\left\{\frac{a_{k+1}}{a_k},\frac{a_{k+2}}{a_{k+1}},...\right\}\\
.\\
.\\
\end{align}$$
Then $\lim\sup\frac{a_{n+1}}{a_n}$ is defined as $\inf\left\{b_1,b_2,...\right\}$. More clearly,
$$\lim\sup\frac{a_{n+1}}{a_n}=\inf_{n\geq1}\left\{\sup_{k\geq n}\left\{\frac{a_{k+1}}{a_k},\frac{a_{k+2}}{a_{k+1}},...\right\}\right\}$$
Now, it's given that $\lim\sup\frac{a_{n+1}}{a_n}=L$. Thus, going by the definition of $\lim\sup$, we can say that, $L=\inf\left\{b_1,b_2,...\right\}$. Also, it has been given that $L'>L$. So can you now apply the definition of the infimum of a set and take it up from here?

 Hint:
$L'$ is not a lower bound of the set $\left\{b_1,b_2,...\right\}$
$\implies\exists r\in\mathbb{N}$ such that $b_r<L'$.
 Now just substitute $b_r=\sup_{k\geq r}\frac{a_{k+1}}{a_k}<L'$. Now as
 the supremum of this whole set is less than $L'$, thus, each and every
 element of this set is less than $L'$. Hence, we found an
 $r\in\mathbb{N}$ such that $\frac{a_{n+1}}{a_n}<L'$ $\forall n\geq r$.

