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Let's say we have an increasing sequence $a_n$ such that $\underset{n\rightarrow\infty}{\lim} a_n=\infty$. Now it's fairly clear to me, though I haven't proven this yet, that: $$\underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=K>0 \Rightarrow \exists\ \delta>0: \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=1$$ In fact, I think $\delta=1/K$ but I could have that mixed up. And as a corrollary: $$\underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=0 \Rightarrow \forall\ \delta>0: \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=0$$

But here's my question: is the converse also true? Specifically is it the case that: $$ \forall\ \delta>0, \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=0 \Rightarrow \underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=0 $$

I think that this is true but when I try to show it, the problem comes up that I can show that $a_n$ is dominated by the same sequence that dominates $\sum\limits_{k=1}^n a_k$, but I can't show that $\sum\limits_{k=1}^n a_k$ is large relative to $a_n$. In my mind though, I know that if the RHS is not true and this limit is positive, then at some point the sequence is proportional to its own rate of change, and that means it should be asymptotically equivalent to an exponential function. But the details of the proof are killing me.

Any help would be greatly appreciated :-)

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    $\begingroup$ Why is it clear to you if you don't have a proof yet? $\endgroup$ – Jef L Apr 17 '15 at 11:16
  • $\begingroup$ In your "fairly clear" property you can't necessarily get the limit to be $1$ -- if you have a sequence where the limit is $1$, you can double every term to get one where the limit is $2$, for example. So the best you can hope for is that the limit exists and is positive. $\endgroup$ – Henning Makholm Apr 17 '15 at 11:22
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Here is a counterexample to your first conjecture, stating that for every unbounded increasing sequence $(a_n)$, $$\underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=K>0 \Rightarrow \exists\ \delta>0: \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=1$$ Take $a_n=n2^n$. Then $\{a_n\}$ is increasing and $a_n\to\infty$. Then $$ a_1+\cdots+a_n=(2-1)(a_1+\cdots+a_n)=2^2+2\cdot 2^3+\cdots+n2^{n+1}-2^1-2\cdot 2^2-\cdots-n2^{n}=n2^{n+1}-2-(2^2+2^3+\cdots+2^n)=n2^{n+1}-2-2^{n+1}+2^2=(n-1)2^{n+1}+2. $$ Also $$ \frac{a_n}{a_1+\cdots+a_n}=\frac{n2^n}{(n-1)2^{n+1}+2}\to \frac{1}{2} $$ But, there is no $\delta>0$, such that $$ \frac{a_n}{\mathrm{e}^{n\delta}}=\frac{n2^n}{\mathrm{e}^{n\delta}}= \exp(\log n+n(\log 2-\delta)) $$ converges to $1$, as $\log n+n(\log 2-\delta)\to \infty$ or $\log n+n(\log 2-\delta)\to -\infty$, if $\delta\ge\log 2$ or $\delta<\log 2$, respectively.

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    $\begingroup$ Sooo... you give an example where the hypothesis is not met and you show the conclusion does not hold. This does not tell much about the validity of the implication, does it? $\endgroup$ – Did Apr 17 '15 at 11:35
  • $\begingroup$ This hypotheses are all met! $\endgroup$ – Yiorgos S. Smyrlis Apr 17 '15 at 11:53
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    $\begingroup$ No. If $a_n=n2^n$, then $a_n/e^{\delta n}$ does not converge to $0$ for every positive $\delta$. $\endgroup$ – Did Apr 17 '15 at 11:58
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    $\begingroup$ I have given a counterexample to the first conjecture! $\endgroup$ – Yiorgos S. Smyrlis Apr 17 '15 at 12:25
  • $\begingroup$ OK, I had missed that. Since this is not the main question the OP is interested in, this should be mentioned explicitely though. I modified your post accordingly. $\endgroup$ – Did Apr 17 '15 at 12:47
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Let $a_k=e^{\beta k}$ with $0<\beta<\delta$ so that your assumption holds. Then $$\sum_1^n a_k=\frac{e^{(n+1)\beta}-e^{\beta}}{e^{\beta}-1}$$ now note that $$\lim_{n \to \infty} \frac{e^{\beta n}}{\frac{e^{(n+1)\beta}-e^{\beta}}{e^{\beta}-1}}=1-e^{-\beta}$$

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Not true.

Take the sequence $\{a_n\}$ of the form: $$ a_n=b_j,\quad \text{for}\,\,\,k_j\le n< k_{j+1}, $$ such that

a. The $b_j$ are positive and form an increasing sequence,

b. $b_j=a_1+\cdots+a_{k_j-1}$, and hence $\dfrac{a_{k_j}}{a_1+\cdots+a_{k_j}}=\dfrac{1}{2}$,

c. $k_j$ is chosen so that $\dfrac{a_{k_j-1}}{a_1+\cdots+a_{k_j-1}}<\dfrac{1}{2^j}$.

d. Such sequence is: $1,1,2,2,2,8,8,8,8,8,8,8,64,\cdots,64,2^{10},\cdots$, which satisfies $$ a_{2^n} \le 2^{n(n+1)/2}\quad \text{and}\quad a_m\le m^{\log m}, $$ and hence $$ 0<\frac{a_m}{\mathrm{e}^{\delta m}}\le\frac{m^{\log m}}{\mathrm{e}^{\delta m}}\to 0, \quad\text{for all $\delta>0$.} $$

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Not true.

Take $a_n = n * (\frac{e}{2})^{n}$. It's easy to verify that $\lim \frac{a_n}{e^n} = 0$

On the other hand, $\frac{a_n}{\sum a_n} = \frac{1}{\sum a_i/a_n}$, where i runs from 1 to n

Now consider the denominator:

$\sum \frac{a_i}{a_n} = \sum \frac{i}{n}*(\frac{e}{2})^{i-n} $ < $\sum (\frac{e}{2})^{i-n}$

Let $k = \frac{e}{2}$. Then, the RHS can be written as: $\frac{1}{k^n}*\sum k^i$ = $\frac{k-1/k^n}{k-1} < \frac{k}{k-1} = \frac{e}{e-2}$

Thus, $\frac{a_n}{\sum a_n} = \frac{1}{\sum a_i/a_n} > \frac{e-2}{e} = constant > 0 $.

Then it can't go to $0$.

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  • $\begingroup$ Can someone help me to find a suitable syntax for the 5th line, where $e/2$ to the power of $(i-n)$ is not correctly displayed? $\endgroup$ – SiXUlm Apr 17 '15 at 12:30
  • $\begingroup$ You need to use {} instead of () $\endgroup$ – Litho Apr 17 '15 at 12:57
  • $\begingroup$ @Litho: it works now, thanks! $\endgroup$ – SiXUlm Apr 17 '15 at 13:09
  • $\begingroup$ Can someone who down voted my answer explain why you down vote? $\endgroup$ – SiXUlm Apr 17 '15 at 13:09

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