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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.

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  • $\begingroup$ Never seen a different proof... $\endgroup$ – Siminore May 3 '14 at 10:54
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    $\begingroup$ The proof invokes at every step about the only sensible next step, so what is awkward? $\endgroup$ – Hagen von Eitzen May 3 '14 at 13:07
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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.

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  • $\begingroup$ Perhaps even more intuitive, if the limsup is less than $1,$ then there is a number $\alpha$ less than $1$ such that every term in some tail of the sequence of ratios is less than or equal to $\alpha.$ (If not, then we can get ratios arbitrarily far out in the sequence that are arbitrarily close to $1,$ and hence a subsequence of ratios converging to $1,$ which contradicts the limsup being less than $1.)$ Now use a direct comparison of the tail with an appropriate geometric series. $\endgroup$ – Dave L. Renfro May 9 '14 at 19:53
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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.

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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, $$\left| \limsup_{m\to\infty}\sum_{k=1}^m |a_k| - \left( \sum_{k=1}^{n} |a_k| + |a_n|\left( \frac {1} {1-r} - 1\right)\right)\right | < |a_n| \lim_{m\to\infty} m r^m = \epsilon $$

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