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Suppose $\displaystyle\lim_{n\to\infty} {\frac{a_{n+1}}{a_n}} = \frac{1}{2}$. Prove $\displaystyle\lim_{n\to\infty}{a_n}=0$.

My start on this was to state the limit in terms of the definition and work on it from there:

$\displaystyle\lim_{n\to\infty} {\frac{a_{n+1}}{a_n}} = \frac{1}{2}$

$\implies \exists n_0 $ s.t., $\forall \epsilon>0, |\frac{a_{n+1}}{an} -\frac{1}{2}| < \epsilon$, $(\forall n_0 > n)$

Using the triangle inequality I restated the inequality above as follows:

$|\frac{a_{n+1}}{an}| \leq |\frac{a_{n+1}}{an} -\frac{1}{2}| + \frac{1}{2} < \epsilon$

$\implies |\frac{a_{n+1}}{an}| < \epsilon + \frac{1}{2}$

$\implies |{a_{n+1}}| < |a_n|(\epsilon + \frac{1}{2})$

At this point I am a bit of a loss. Am I permitted to generalize the above result as:

$\implies |{a_{n+k}}| < |a_n|(\epsilon + \frac{1}{2})^k$ ?

And if so, is there some way I can utilize this fact to arrive at a proof of $a_n \rightarrow 0$ ?

Thank you in advance for any help. :)

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  • $\begingroup$ hint: limit of $a_n$ is $k\left(\frac{1}{2}\right)^n$ $\endgroup$ – Saketh Malyala Feb 2 '17 at 4:09
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You're on the right track, now WLOG assume $\epsilon < {1\over 2}$ so that after you reach the appropriate $N\in\Bbb N$ when your hypothesis starts to hold, with $0<a=\epsilon + {1\over 2} < 1$ and then you have $|a_{N+k}|<a^k|a_N|\to 0$ as $k\to\infty$ proving the result. You can even quantify it if you like, since

$$a^k|a_N|<\epsilon\iff k>{\log\epsilon -\log|a_N|\over\log a}$$

where the inequality switches direction because we divide by the negative number $\log a$.

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  • $\begingroup$ Thank you for the answer Adam. Sorry if this sounds a bit dense , but trying to make sense of this, does this result "work" because $a_{N+k}$ is essentially the same sequence as $a_N$ and thus if we can prove that $a_{N+k} \rightarrow 0$ then we have shown that $a_{N} \rightarrow 0$? And is this because (if I interpret what you said correctly) because $\forall \epsilon > 0$ we can find a sufficiently large $N \in \mathbb{N}$ s.t. $a^k|a_N| < \epsilon$? Thank you again :) $\endgroup$ – Fernando Villasenor Feb 2 '17 at 3:54
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    $\begingroup$ @FernandoVillasenor yes, that's correct, because it's just a shifted sequence, you can let $n=N+k$ and then $n\to\infty\iff k\to\infty$ since $N$ is fixed. $\endgroup$ – Adam Hughes Feb 2 '17 at 4:12
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You can find some n so that all the ratios are under 3/4 (e = 1/4) so the limit is less than (a sub n) × (3/4) ^ m as m goes to infinity.

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  • $\begingroup$ Can you elaborate a little more on why you are choosing 3/4 specifically (or in other words why you have chosen $\epsilon = \frac{1}{4} $) $\endgroup$ – Fernando Villasenor Feb 2 '17 at 6:17
  • $\begingroup$ I just picked some number that would keep the ratio under 1. Since the sequence converges to 0, you can definitely pick 1/4 as a number. $\endgroup$ – Mike O'Connor Feb 3 '17 at 3:57
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More generally, if each $a_n\ne 0$ and $1>V=\lim_{n\to \infty}\sup_{m\geq n}|\frac {a_{m+1}}{a_m}|$ then $\lim_{n\to \infty}a_n=0.$

Proof: Let $n_0$ be such that $\forall n\geq n_0\left(\; |\frac {a_{n+1}}{a_n}|\leq \frac {1+V}{2}\;\right).$ Now for $n\geq n_0$ we have $$|a_{n+1}|=|a_{n_0}|\cdot \prod_{j=n_0}^n\left|\frac {a_{j+1}}{a_j}\right|\leq |a_{n_0}|\left(\frac {1+V}{2}\right)^{n-n_0+1}.$$ The far RHS expression goes to $0$ as $n\to \infty$ because $0<\frac {1+V}{2}<1.$

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