Limsup Proof; Is there a more intuitive proof? 
Suppose that $(a_{n}:n\geq1)$ is a sequence satisfying $\limsup_{n \to \infty}|\dfrac{a_{n+1}}{a_n}|<1$.
  Prove that $\sum_{n=0}^{\infty}a_n$ converges absolutly.  

Now the proof given is a bit awkward is there a nicer more intuitive proof than this ;  

If $\limsup_{n \to \infty}b_n=\alpha$ then for any $\epsilon>0 \exists$ only finitely many $n$ such that $b_n\geq \alpha + \epsilon$ .
  So if $\limsup_{n \to \infty}|\dfrac{a_{n+1}}{a_n}|<\alpha + \epsilon$ we can pick $\epsilon$ such that $\alpha +\epsilon < 1$ and so $\exists N_0$ such that $|\dfrac{a_{n+1}}{a_n}|<\alpha +\epsilon$ for all $n\geq N_0$.
  So    $|a_{N_{0}+m}|\leq(\alpha +\epsilon)|a_{N_{0}+m-1}|\leq (\alpha +\epsilon)^{n}|a_{N_{0}}|$  $\forall m\geq1$.
  Consequently there exists a $B$ such that $|a_n|\leq B(\alpha +\epsilon)^{n}$ for all n and by comparison $\sum_{n=0}^{\infty}|a_n|$ is convergent.

 A: The proof is actually quite intuitive, I think, but the presentation included in your answer fails to emphasize that. The idea is simple, really - we compare to a geometric series. We start from $$
  \limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \alpha < 1 \text{,}
$$
and from the knowledge that the geometric series $$
  \sum_{k=0}^\infty c\beta^k
$$
converges if $0 < \beta < 1$.
Now, instead of $\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \alpha < 1$ we had $\left|\frac{a_{n+1}}{a_n}\right| < \alpha < 1$ for all $n$, it'd follow that $|a_n| < |a_0|\alpha^n$ and we'd be done, by comparison to the geometric series above (set $\beta = \alpha$ and $c=|a_0|$).
Now, we don't quite have that, but we have something close enough. By the very definition of $\limsup$, we know that for every $\epsilon > 0$,  $\left|\frac{a_{n+1}}{a_n}\right| > \alpha + \epsilon$ can occur only finitely many times. So what we do is we pick an $\epsilon$ such that $$
  \alpha + \epsilon < 1\text{, say $\epsilon = \frac{1 - \alpha}{2}$, and set $\beta = \alpha + \epsilon$.}
$$
Since $\left|\frac{a_{n+1}}{a_n}\right| > \beta$ occurs only finitely many times, there's an $N$ such that $$
  \left|\frac{a_{n+1}}{a_n}\right| < \beta \text{ if } n \geq N \text{,}
$$
and therefore $$
  |a_n| \leq |a_N|\beta^{n-N} \text{ if $n \geq N$.}
$$
But then were' done! We just have to split the sum into two parts - the terms up to $a_N$ and those following $a_N$ - and compare the latter with the geometric series to get $$
  \sum_{n=0}^\infty |a_n| = \underbrace{\sum_{n=0}^{N-1} |a_n|}_{\text{$< \infty$ since finite sum}} + \underbrace{\sum_{n=N}^\infty |a_n|}_{\leq \sum_{k=0}^\infty |a_N|\beta^k < \infty} < \infty \text{.}
$$
Your proof does exactly the same, just in much fewer words.
A: The intuition of your proof is the standard one for this theorem. Nevertheless it can be said that introducing $\alpha$ and $\epsilon$ without reference to the actually needed values is not optimal. And read again this sentence of yours: "So if $\limsup_{n \to \infty}|\dfrac{a_{n+1}}{a_n}|<\alpha + \epsilon$ we can pick $\epsilon$ such that $\alpha +\epsilon < 1$."
Having said this, I'd argue as follows:
Assume $\limsup_{n\to\infty}\left|{a_{n+1}\over a_n}\right|=:p<1$ and choose a $q$ with $p<q<1$. Then there is an $n_0$ with $\left|{a_{n+1}\over a_n}\right|\leq q$ for all $n\geq n_0$, so that by induction we obtain
$$|a_n|\leq q^n {|a_{n_0}|\over q^{n_0}}\qquad(n\geq n_0)\ .$$
This ensures the absolute convergence of $\sum_{n=0}^\infty a_n$ by the comparison test.
A: Just an attempt. Not sure, if incorrect please suggest.
Let $\limsup_{n \to \infty}\left|\dfrac{a_{n+1}}{a_n}\right| =  r < 1 $, then $ \exists n \ge N \in  \Bbb N  $ such that $\forall n' > n (||a_{n'}| - r |a_{n'+1}|| < \delta (n)  = r^n )$.
After which we get, 
$$\limsup_{m\to\infty}\sum_{k=1}^m |a_k|  <   \sum_{k=1}^{n} |a_k| + |a_n| \lim_{m\to\infty} ( r + r^2++ \dots +r^{m-1} + n \delta(n)) \\\le  \sum_{k=1}^{n} |a_k| + |a_n|\left( \frac {1} {1-r}   - 1\right) + m \delta(m) |a_n|$$
Or, 
$$\left| \limsup_{m\to\infty}\sum_{k=1}^m |a_k| - \left(  \sum_{k=1}^{n} |a_k| + |a_n|\left( \frac {1} {1-r}   - 1\right)\right)\right | < |a_n| \lim_{m\to\infty} m r^m = \epsilon $$
